Let $\hat { a }$ be a unit vector perpendicular to the vector $\overrightarrow { \mathrm { b } } = \hat { i } - 2 \hat { j } + 3 \hat { k }$ and $\overrightarrow { \mathrm { c } } = 2 \hat { i } + 3 \hat { j } - \hat { k }$, and makes an angle of $\cos ^ { - 1 } \left( - \frac { 1 } { 3 } \right)$ with the vector $\hat { i } + \hat { j } + \hat { k }$. If $\hat { a }$ makes an angle of $\frac { \pi } { 3 }$ with the vector $\hat { i } + \alpha \hat { j } + \hat { k }$, then the value of $\alpha$ is :
(1) $\sqrt { 6 }$
(2) $- \sqrt { 6 }$
(3) $- \sqrt { 3 }$
(4) $\sqrt { 3 }$

Let the position vectors of the vertices $A , B$ and $C$ of a tetrahedron $A B C D$ be $\hat { \mathbf { i } } + 2 \hat { \mathbf { j } } + \hat { \mathbf { k } } , \hat { \mathbf { i } } + 3 \hat { \mathbf { j } } - 2 \hat { k }$ and $2 \hat { i } + \hat { j } - \hat { k }$ respectively. The altitude from the vertex $D$ to the opposite face $A B C$ meets the median line segment through $A$ of the triangle $A B C$ at the point $E$. If the length of $A D$ is $\frac { \sqrt { } \overline { 110 } } { 3 }$ and the volume of the tetrahedron is $\frac { \sqrt { 805 } } { 6 \sqrt { 2 } }$, then the position vector of $E$ is
(1) $\frac { 1 } { 12 } ( 7 \hat { \mathbf { i } } + 4 \hat { \mathbf { j } } + 3 \hat { k } )$
(2) $\frac { 1 } { 2 } ( \hat { i } + 4 \hat { j } + 7 \hat { k } )$
(3) $\frac { 1 } { 6 } ( 12 \hat { i } + 12 \hat { j } + \hat { k } )$
(4) $\frac { 1 } { 6 } ( 7 \hat { \mathrm { i } } + 12 \hat { \mathrm { j } } + \hat { \mathrm { k } } )$

Let a line pass through two distinct points $P ( - 2 , - 1,3 )$ and $Q$, and be parallel to the vector $3 \hat { i } + 2 \hat { j } + 2 \hat { k }$. If the distance of the point Q from the point $\mathrm { R } ( 1,3,3 )$ is 5 , then the square of the area of $\triangle P Q R$ is equal to :
(1) 148
(2) 136
(3) 144
(4) 140

Let P be the foot of the perpendicular from the point $( 1,2,2 )$ on the line $\mathrm { L } : \frac { x - 1 } { 1 } = \frac { y + 1 } { - 1 } = \frac { z - 2 } { 2 }$. Let the line $\vec { r } = ( - \hat { i } + \hat { j } - 2 \hat { k } ) + \lambda ( \hat { i } - \hat { j } + \hat { k } ) , \lambda \in \mathbf { R }$, intersect the line L at Q . Then $2 ( \mathrm { PQ } ) ^ { 2 }$ is equal to :
(1) 25
(2) 19
(3) 29
(4) 27

If the square of the shortest distance between the lines $\frac { x - 2 } { 1 } = \frac { y - 1 } { 2 } = \frac { z + 3 } { - 3 }$ and $\frac { x + 1 } { 2 } = \frac { y + 3 } { 4 } = \frac { z + 5 } { - 5 }$ is $\frac { \mathrm { m } } { \mathrm { n } }$, where $\mathrm { m } , \mathrm { n }$ are coprime numbers, then $\mathrm { m } + \mathrm { n }$ is equal to :
(1) 21
(2) 9
(3) 14
(4) 6

Let the line passing through the points $(-1, 2, 1)$ and parallel to the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z}{4}$ intersect the line $\frac{x+2}{3} = \frac{y-3}{2} = \frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4, -5, 1)$ is
(1) 5
(2) $5\sqrt{5}$
(3) $5\sqrt{6}$
(4) 10

Let $\overrightarrow{\mathrm{a}} = \hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{\mathrm{b}} = 2\hat{i} + 7\hat{j} + 3\hat{k}$. Let $\mathrm{L}_1: \overrightarrow{\mathrm{r}} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda\overrightarrow{\mathrm{a}},\ \lambda \in \mathbf{R}$ and $\mathrm{L}_2: \overrightarrow{\mathrm{r}} = (\hat{j} + \hat{k}) + \mu\overrightarrow{\mathrm{b}},\ \mu \in \mathbf{R}$ be two lines. If the line $\mathrm{L}_3$ passes through the point of intersection of $\mathrm{L}_1$ and $\mathrm{L}_2$, and is parallel to $\vec{a} + \vec{b}$, then $\mathrm{L}_3$ passes through the point:
(1) $(5, 17, 4)$
(2) $(2, 8, 5)$
(3) $(8, 26, 12)$
(4) $(-1, -1, 1)$

Let $P$ be the foot of the perpendicular from the point $Q ( 10 , - 3 , - 1 )$ on the line $\frac { x - 3 } { 7 } = \frac { y - 2 } { - 1 } = \frac { z + 1 } { - 2 }$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $( 3 , - 2,1 )$, is
(1) $9 \sqrt { 15 }$
(2) $\sqrt { 30 }$
(3) $8 \sqrt { 15 }$
(4) $3 \sqrt { 30 }$

If the image of the point $( 4,4,3 )$ in the line $\frac { x - 1 } { 2 } = \frac { y - 2 } { 1 } = \frac { z - 1 } { 3 }$ is $( \alpha , \beta , \gamma )$, then $\alpha + \beta + \gamma$ is equal to
(1) 9
(2) 12
(3) 7
(4) 8

Let the position vectors of three vertices of a triangle be $4\vec{p} + \vec{q} - 3\vec{r}$, $-5\vec{p} + \vec{q} + 2\vec{r}$ and $2\vec{p} - \vec{q} + 2\vec{r}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p} + \vec{q} + \vec{r}}{4}$ and $\alpha\vec{p} + \beta\vec{q} + \gamma\vec{r}$ respectively, then $\alpha + 2\beta + 5\gamma$ is equal to:
(1) 3
(2) 4
(3) 1
(4) 6

Let $\mathrm { A } ( x , y , z )$ be a point in $xy$-plane, which is equidistant from three points $( 0,3,2 ) , ( 2,0,3 )$ and $( 0,0,1 )$. Let $\mathrm { B } = ( 1,4 , - 1 )$ and $\mathrm { C } = ( 2,0 , - 2 )$. Then among the statements (S1) : $\triangle \mathrm { ABC }$ is an isosceles right angled triangle, and (S2) : the area of $\triangle \mathrm { ABC }$ is $\frac { 9 \sqrt { 2 } } { 2 }$,
(1) both are true
(2) only (S2) is true
(3) only (S1) is true
(4) both are false

Let $\vec{\mathrm{a}} = 3\hat{i} - \hat{j} + 2\hat{k}$, $\vec{\mathrm{b}} = \vec{\mathrm{a}} \times (\hat{i} - 2\hat{k})$ and $\vec{\mathrm{c}} = \vec{\mathrm{b}} \times \hat{k}$. Then the projection of $\vec{\mathrm{c}} - 2\hat{j}$ on $\vec{a}$ is:
(1) $2\sqrt{14}$
(2) $\sqrt{14}$
(3) $3\sqrt{7}$
(4) $2\sqrt{7}$

Let $\mathrm{L}_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}$ and $\mathrm{L}_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $\mathrm{L}_1$, then $|5\alpha - 11\beta - 8\gamma|$ equals:
(1) 20
(2) 18
(3) 25
(4) 16

The perpendicular distance, of the line $\frac { x - 1 } { 2 } = \frac { y + 2 } { - 1 } = \frac { z + 3 } { 2 }$ from the point $\mathrm { P } ( 2 , - 10,1 )$, is :
(1) 6
(2) $5 \sqrt { 2 }$
(3) $4 \sqrt { 3 }$
(4) $3 \sqrt { 5 }$

Let a straight line $L$ pass through the point $P ( 2 , - 1,3 )$ and be perpendicular to the lines $\frac { x - 1 } { 2 } = \frac { y + 1 } { 1 } = \frac { z - 3 } { - 2 }$ and $\frac { x - 3 } { 1 } = \frac { y - 2 } { 3 } = \frac { z + 2 } { 4 }$. If the line $L$ intersects the $y z$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is :
(1) $\sqrt { 10 }$
(2) $2 \sqrt { 3 }$
(3) 2
(4) 3

The square of the distance of the point $\left( \frac { 15 } { 7 } , \frac { 32 } { 7 } , 7 \right)$ from the line $\frac { x + 1 } { 3 } = \frac { y + 3 } { 5 } = \frac { z + 5 } { 7 }$ in the direction of the vector $\hat { i } + 4 \hat { j } + 7 \hat { k }$ is :
(1) 54
(2) 44
(3) 41
(4) 66

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c} = 5$, then $|\vec{c}|$ is equal to
(1) $\sqrt{\frac{11}{6}}$
(2) $\frac{1}{3\sqrt{2}}$
(3) 16
(4) 18

Let $\mathrm { L } _ { 1 } : \frac { x - 1 } { 3 } = \frac { y - 1 } { - 1 } = \frac { z + 1 } { 0 }$ and $\mathrm { L } _ { 2 } : \frac { x - 2 } { 2 } = \frac { y } { 0 } = \frac { z + 4 } { \alpha } , \alpha \in \mathbf { R }$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A ( 1,1 , - 1 )$ on $L _ { 2 }$, then the value of $26 \alpha ( \mathrm {~PB} ) ^ { 2 }$ is $\_\_\_\_$

Suppose that a triangle ABC which is inscribed in a circle O of radius 2 satisfies
$$3\overrightarrow{\mathrm{OA}} + 4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}} = \vec{0}.$$
Let D denote the point of intersection of the straight line AO and the segment BC. We are to find the lengths of the segments AD and BD.
(1) When we set $\overrightarrow{\mathrm{OD}} = k\overrightarrow{\mathrm{OA}}$ where $k$ is a real number, we have
$$\overrightarrow{\mathrm{OD}} = -\frac{\mathbf{A}}{\mathbf{B}}k\overrightarrow{\mathrm{OB}} - \frac{\mathbf{C}}{\mathbf{D}}k\overrightarrow{\mathrm{OC}}.$$
As the three points B, C and D are located on a straight line, we obtain $k = \frac{\mathbf{EF}}{\mathbf{GH}}$. From this we derive that $\mathrm{OD} = \mathbf{H}$ and finally obtain
$$\mathrm{AD} = 1.$$
(2) From (1) we see that $\mathrm{BD} = \frac{\mathbf{J}}{\mathbf{K}}\mathrm{BC}$. So in order to find the length of the segment BD, we should find the length of the segment BC.
First we note that
$$\mathrm{BC}^2 = \mathbf{L} - \mathbf{M}\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$$
where $\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$ represents the inner product of $\overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{OC}}$. Since we know from (1) that $|4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}}|^2 = \mathbf{NO}$, we have
$$\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}} = \frac{\mathbf{PQR}}{\mathbf{S}}.$$
Hence we obtain $\mathrm{BC} = \frac{\square\sqrt{\mathbf{U}}}{\square\mathbf{V}}$ and finally from that
$$\mathrm{BD} = \frac{\sqrt{\mathrm{W}}}{\mathrm{X}}.$$

Given a sphere $S$ whose center is at O and whose radius is 1 , take three points A , B and C on $S$ such that
$$\overrightarrow { \mathrm { OA } } \cdot \overrightarrow { \mathrm { OB } } = \overrightarrow { \mathrm { OB } } \cdot \overrightarrow { \mathrm { OC } } = \overrightarrow { \mathrm { OC } } \cdot \overrightarrow { \mathrm { OA } } = 0 .$$
Note that $\overrightarrow { \mathrm { OA } } \cdot \overrightarrow { \mathrm { OB } }$, etc., refers to the inner product of the two vectors.
(1) It follows that $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { AC } } = \square \mathbf { A } , | \overrightarrow { \mathrm { AB } } | = \sqrt { \mathbf { B } } , \cos \angle \mathrm { BAC } = \frac { \square \mathbf { C } } { \square }$ and the area of the triangle ABC is $\frac { \sqrt { \mathbf { E } } } { \mathbf { F } }$.
(2) Let G be the center of gravity of triangle ABC and P be the intersection point of the ray (half line) OG and sphere $S$.
Since $\overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { G } } ( \overrightarrow { \mathrm { OA } } + \overrightarrow { \mathrm { OB } } + \overrightarrow { \mathrm { OC } } )$, we have
$$\begin{gathered} | \overrightarrow { \mathrm { OG } } | = \frac { \sqrt { \mathbf { I } } } { \sqrt { \mathbf { J } } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \square \mathbf { K } - \sqrt { \square \mathbf { L } } } { \square \mathbf { M } } \\ \overrightarrow { \mathrm { AG } } \cdot \overrightarrow { \mathrm { PG } } = \square \mathbf { N } . \end{gathered}$$
Hence the volume of the tetrahedron PABC is $\frac { \sqrt { \mathbf { Q } } - \mathbf { P } } { \mathbf { Q } }$.

In a regular tetrahedron OABC, each side of which has the length 1, let L denote the point which divides segment OA internally in the ratio $3:1$, let M denote the midpoint of side BC, and let P denote the point which divides segment LM internally in the ratio $t:(1-t)$, where $0 < t < 1$.
(1) When we set $\overrightarrow{\mathrm{OA}} = \vec{a}$, $\overrightarrow{\mathrm{OB}} = \vec{b}$, $\overrightarrow{\mathrm{OC}} = \vec{c}$ and express $\overrightarrow{\mathrm{OP}}$ in terms of $\vec{a}, \vec{b}, \vec{c}$, we have
$$\overrightarrow{\mathrm{OP}} = \frac{\mathbf{A}}{\mathbf{B}}(\mathbf{C} - t)\vec{a} + \frac{\mathbf{D}}{\mathbf{E}}t(\vec{b} + \vec{c}).$$
Since $\vec{a} \cdot (\vec{b} + \vec{c}) = \mathbf{F}$ and $|\vec{b} + \vec{c}|^2 = \mathbf{G}$, we have
$$|\overrightarrow{\mathrm{OP}}| = \frac{1}{\mathbf{H}}\sqrt{\mathbf{I}t^2 - \mathbf{JJ}t + \mathbf{K}},$$
where $\vec{a} \cdot (\vec{b} + \vec{c})$ means the inner product of the vectors $\vec{a}$ and $(\vec{b} + \vec{c})$.
(2) The value of $t$ at which $|\overrightarrow{\mathrm{OP}}|$ is minimized is
$$t = \frac{\mathbf{L}}{\mathbf{L}}$$
and the minimum value of $|\overrightarrow{\mathrm{OP}}|$ is $\frac{\sqrt{\mathbf{N}}}{\mathbf{O}}$.
(3) When $|\overrightarrow{\mathrm{OP}}|$ is minimized as in (2), we have $\cos\angle\mathrm{AOP} = \frac{\square\mathbf{P}\sqrt{\mathbf{Q}}}{\square\mathbf{R}}$.

Let $S$ be a circle with its center at point O and a radius of 1. Let $\triangle \mathrm { ABC }$ be a triangle such that all its vertices are on $S$ and $\mathrm { AB } : \mathrm { AC } = 3 : 2$. As shown in the figure, let D be a point on the extension of side BC and $k$ be the number where
$$\mathrm { BC } : \mathrm { CD } = 2 : k .$$
Moreover, set
$$\overrightarrow { \mathrm { OA } } = \vec { a } , \quad \overrightarrow { \mathrm { OB } } = \vec { b } , \quad \overrightarrow { \mathrm { OC } } = \vec { c }$$
Answer the following questions.
(1) When we express $\overrightarrow { \mathrm { OD } }$ in terms of $\vec { b } , \vec { c }$ and $k$, we have
$$\overrightarrow { \mathrm { OD } } = \left( \frac { k } { \mathbf { A } } + \mathbf { B } \right) \vec { c } - \frac { k } { \mathbf { C } } \vec { b }$$
(2) Since the equality
$$| \vec { b } - \vec { a } | = \frac { \mathbf { D } } { \mathbf { E } } | \vec { c } - \vec { a } |$$
holds, by expressing the inner product $\vec { a } \cdot \vec { b }$ in terms of the inner product $\vec { a } \cdot \vec { c }$, we have
$$\vec { a } \cdot \vec { b } = \frac { \mathbf { F } } { \mathbf { G } } \vec { a } \cdot \vec { c } - \frac { \mathbf { H } } { \mathbf { I } }$$
(3) It follows that when the tangent to $S$ at the point A passes through the point D,
$$k = \frac { \mathbf { J } } { \mathbf { K } } .$$

Given two points $\mathrm { A } ( 1 , - 1,0 )$ and $\mathrm { B } ( - 2,1,2 )$ in a coordinate space with the origin O, let us set $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$.
(1) First, we are to find the value of $t$ at which $| \vec { a } + t \vec { b } |$ is minimized. Since
$$| \vec { a } + t \vec { b } | ^ { 2 } = \mathbf { A } t ^ { 2 } - \mathbf { B } t + \mathbf { C }$$
$| \vec { a } + t \vec { b } |$ is minimized at $t = \frac { \mathbf { D } } { \mathbf { E } }$, and its minimum value is $\mathbf { F }$.
(2) Next, the vectors $\vec { c }$ which are orthogonal to the vectors $\vec { a }$ and $\vec { b }$ can be represented as
$$\vec { c } = s ( \mathbf { G } , \mathbf { H } , 1 )$$
where $s$ is a non-zero real number. Now, let C and D be the points such that $\overrightarrow { \mathrm { OC } } = ( \mathbf { G } , \mathbf { H } , 1 )$ and $\overrightarrow { \mathrm { OD } } = 3 \vec { a } + \vec { b }$. Since $\angle \mathrm { CBD } = \frac { \pi } { \mathbf { I } }$, the area of the triangle BCD is $\frac { \mathbf { J } \sqrt { \mathbf { K } } } { \mathbf { L } }$.

For $\mathbf { C } , \mathbf { D } , \mathbf { E } , \mathbf { F } , \mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below. For the other $\square$, enter the correct number.
Consider a regular tetrahedron OABC with sides of length 1. Let $x$ be a number satisfying $0 < x < 1$, and let P be the point that divides side AB by the ratio $x : ( 1 - x )$ and Q be the point that divides side BC by the ratio $x : ( 1 - x )$. Also, let $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$. We are to find the range of values of $\cos \angle \mathrm { POQ }$.
The vectors $\vec { a } , \vec { b }$ and $\vec { c }$ satisfy
$$\vec { a } \cdot \vec { b } = \vec { b } \cdot \vec { c } = \vec { c } \cdot \vec { a } = \frac { \mathbf { A } } { \mathbf { B } }$$
Next, since we can express $\overrightarrow { \mathrm { OP } }$ and $\overrightarrow { \mathrm { OQ } }$ as $\overrightarrow { \mathrm { OP } } = \square \mathbf { C }$ and $\overrightarrow { \mathrm { OQ } } = \mathbf { D }$, we have
$$| \overrightarrow { \mathrm { OP } } | = | \overrightarrow { \mathrm { OQ } } | = \sqrt { \vec { E } } , \quad \overrightarrow { \mathrm { OP } } \cdot \overrightarrow { \mathrm { OQ } } = \square \mathbf { F } .$$
Hence we obtain
$$\cos \angle \mathrm { POQ } = \frac { 1 } { \mathbf { G } } - \frac { \mathbf { H } } { \mathbf { I } } .$$
From this we finally obtain
$$\frac { \square \mathbf { J } } { \mathbf { K } } < \cos \angle \mathrm { POQ } \leqq \frac { \mathbf { L } } { \mathbf { M } } .$$
(0) $( 1 - x ) \vec { a } + x \vec { b }$
(1) $x \vec { a } + ( 1 - x ) \vec { b }$
(2) $( 1 - x ) \vec { b } + x \vec { c }$
(3) $x \vec { b } + ( 1 - x ) \vec { c }$
(4) $x ^ { 2 } + x + 1$
(5) $x ^ { 2 } - x + 1$ (6) $x ^ { 2 } - x - 1$ (7) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x + 1 \right)$ (8) $\frac { 1 } { 2 } \left( - x ^ { 2 } - x + 1 \right)$ (9) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x - 1 \right)$

In coordinate space, let $O$ be the origin, and let point $P$ be the intersection of three planes $x - 3y - 5z = 0$, $x - 3y + 2z = 0$, $x + y = t$, where $t > 0$. If $\overline{OP} = 10$, then $t =$ （9）（10）（11）. (Express as a simplified radical)