The figure below represents a cube ABCDEFGH with the plane (IJK) having Cartesian equation $4x - 6y - 4z + 3 = 0$ in the orthonormal coordinate system $(\mathrm{A}; \overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AD}}, \overrightarrow{\mathrm{AE}})$. We denote by R the orthogonal projection of point F onto the plane (IJK). Point R is therefore the unique point of the plane (IJK) such that the line (FR) is orthogonal to the plane (IJK). We define the interior of the cube as the set of points $M(x; y; z)$ such that $\left\{\begin{array}{l} 0 < x < 1 \\ 0 < y < 1 \\ 0 < z < 1 \end{array}\right.$ Is point R inside the cube?

In an orthonormal coordinate system $(\mathrm{O}; \vec{\imath}, \vec{\jmath}, \vec{k})$ we consider the points: $$\mathrm{A}(1;1;-4), \quad \mathrm{B}(2;-1;-3), \quad \mathrm{C}(0;-1;-1) \text{ and } \Omega(1;1;2).$$

1. Prove that the points $\mathrm{A}$, $\mathrm{B}$ and C define a plane.
2. a. Prove that the vector $\vec{n}$ with coordinates $\left(\begin{array}{l}1\\1\\1\end{array}\right)$ is normal to the plane (ABC). b. Justify that a Cartesian equation of the plane (ABC) is $x + y + z + 2 = 0$.
3. a. Justify that the point $\Omega$ does not belong to the plane (ABC). b. Determine the coordinates of the point H, the orthogonal projection of the point $\Omega$ onto the plane (ABC).

We admit that $\Omega\mathrm{H} = 2\sqrt{3}$. We define the sphere $S$ with centre $\Omega$ and radius $2\sqrt{3}$ as the set of all points M in space such that $\Omega\mathrm{M} = 2\sqrt{3}$.
4. Justify, without calculation, that any point N of the plane (ABC), distinct from H, does not belong to the sphere $S$. We say that a plane $\mathscr{P}$ is tangent to the sphere $S$ at a point K when the following two conditions are satisfied:

• $\mathrm{K} \in \mathscr{P} \cap S$
• $(\Omega\mathrm{K}) \perp \mathscr{P}$

1. Let the plane $\mathscr{P}$ with Cartesian equation $x + y - z - 6 = 0$ and the point K with coordinates $\mathrm{K}(3;3;0)$. Prove that the plane $\mathscr{P}$ is tangent to the sphere $S$ at point K.
2. We admit that the planes (ABC) and $\mathscr{P}$ intersect along a line ($\Delta$). Determine a parametric equation of the line ($\Delta$).

For a tetrahedron ABCD with an equilateral triangle BCD of side length 12 as one face, let H be the foot of the perpendicular from vertex A to plane BCD. The point H lies inside triangle BCD. The area of triangle CDH is 3 times the area of triangle BCH, the area of triangle DBH is 2 times the area of triangle BCH, and $\overline { \mathrm { AH } } = 3$. Let M be the midpoint of segment BD, and let Q be the foot of the perpendicular from point A to segment CM. What is the length of segment AQ? [4 points]
(1) $\sqrt { 11 }$
(2) $2 \sqrt { 3 }$
(3) $\sqrt { 13 }$
(4) $\sqrt { 14 }$
(5) $\sqrt { 15 }$

As shown in the figure, there is a rhombus-shaped piece of paper ABCD with side length 4 and $\angle \mathrm { BAD } = \frac { \pi } { 3 }$. Let M and N be the midpoints of sides BC and CD respectively. The paper is folded along the three line segments $\mathrm { AM } , \mathrm { AN } , \mathrm { MN }$ to form a tetrahedron PAMN. The area of the orthogonal projection of triangle AMN onto the plane PAM is $\frac { q } { p } \sqrt { 3 }$. Find the value of $p + q$. (Here, the thickness of the paper is neglected, P is the point where the three points $\mathrm { B } , \mathrm { C } , \mathrm { D }$ coincide when the paper is folded, and $p$ and $q$ are coprime natural numbers.) [4 points]

If the distance of the point $\mathrm { P } ( 1 , - 2,1 )$ from the plane $\mathrm { x } + 2 \mathrm { y } - 2 z = \alpha$, where $\alpha > 0$, is 5 , then the foot of the perpendicular from $P$ to the plane is
A) $\left( \frac { 8 } { 3 } , \frac { 4 } { 3 } , - \frac { 7 } { 3 } \right)$
B) $\left( \frac { 4 } { 3 } , - \frac { 4 } { 3 } , \frac { 1 } { 3 } \right)$
C) $\left( \frac { 1 } { 3 } , \frac { 2 } { 3 } , \frac { 10 } { 3 } \right)$
D) $\left( \frac { 2 } { 3 } , - \frac { 1 } { 3 } , \frac { 5 } { 2 } \right)$

Perpendiculars are drawn from points on the line $\frac { x + 2 } { 2 } = \frac { y + 1 } { - 1 } = \frac { z } { 3 }$ to the plane $x + y + z = 3$. The feet of perpendiculars lie on the line
(A) $\frac { x } { 5 } = \frac { y - 1 } { 8 } = \frac { z - 2 } { - 13 }$
(B) $\frac { x } { 2 } = \frac { y - 1 } { 3 } = \frac { z - 2 } { - 5 }$
(C) $\frac { x } { 4 } = \frac { y - 1 } { 3 } = \frac { z - 2 } { - 7 }$
(D) $\frac { x } { 2 } = \frac { y - 1 } { - 7 } = \frac { z - 2 } { 5 }$

In $\mathbb { R } ^ { 3 }$, let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P _ { 1 } : x + 2 y - z + 1 = 0$ and $P _ { 2 } : 2 x - y + z - 1 = 0$. Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P _ { 1 }$. Which of the following points lie(s) on $M$?
(A) $\left( 0 , - \frac { 5 } { 6 } , - \frac { 2 } { 3 } \right)$
(B) $\left( - \frac { 1 } { 6 } , - \frac { 1 } { 3 } , \frac { 1 } { 6 } \right)$
(C) $\left( - \frac { 5 } { 6 } , 0 , \frac { 1 } { 6 } \right)$
(D) $\left( - \frac { 1 } { 3 } , 0 , \frac { 2 } { 3 } \right)$

Let $P$ be a point in the first octant, whose image $Q$ in the plane $x + y = 3$ (that is, the line segment $P Q$ is perpendicular to the plane $x + y = 3$ and the mid-point of $P Q$ lies in the plane $x + y = 3$ ) lies on the $z$-axis. Let the distance of $P$ from the $x$-axis be 5 . If $R$ is the image of $P$ in the $x y$-plane, then the length of $P R$ is $\_\_\_\_$ .

Let $S$ be the reflection of a point $Q$ with respect to the plane given by
$$\vec { r } = - ( t + p ) \hat { \imath } + t \hat { \jmath } + ( 1 + p ) \hat { k }$$
where $t , p$ are real parameters and $\hat { \imath } , \hat { \jmath } , \hat { k }$ are the unit vectors along the three positive coordinate axes. If the position vectors of $Q$ and $S$ are $10 \hat { \imath } + 15 \hat { \jmath } + 20 \hat { k }$ and $\alpha \hat { \imath } + \beta \hat { \jmath } + \gamma \hat { k }$ respectively, then which of the following is/are TRUE ?
(A) $3 ( \alpha + \beta ) = - 101$
(B) $3 ( \beta + \gamma ) = - 71$
(C) $3 ( \gamma + \alpha ) = - 86$
(D) $3 ( \alpha + \beta + \gamma ) = - 121$

If the image of the point $P ( 1 , - 2 , 3 )$ in the plane $2 x + 3 y - 4 z + 22 = 0$ measured parallel to the line $\frac { x } { 1 } = \frac { y } { 4 } = \frac { z } { 5 }$ is $Q$, then $P Q$ is equal to:
(1) $3 \sqrt { 5 }$
(2) $2 \sqrt { 42 }$
(3) $\sqrt { 42 }$
(4) $6 \sqrt { 5 }$

The coordinates of the foot of the perpendicular from the point $( 1 , - 2,1 )$ on the plane containing the lines $\frac { x + 1 } { 6 } = \frac { y - 1 } { 7 } = \frac { z - 3 } { 8 }$ and $\frac { x - 1 } { 3 } = \frac { y - 2 } { 5 } = \frac { z - 3 } { 7 }$, is:
(1) $( 2 , - 4,2 )$
(2) $( 1,1,1 )$
(3) $( 0,0,0 )$
(4) $( - 1,2 , - 1 )$

Let $P$ be a plane passing through the points $(2,1,0)$, $(4,1,1)$ and $(5,0,1)$ and $R$ be any point $(2,1,6)$. Then the image of $R$ in the plane $P$ is
(1) $(6,5,2)$
(2) $(6,5,-2)$
(3) $(4,3,2)$
(4) $(3,4,-2)$

The foot of the perpendicular drawn from the point $( 4,2,3 )$ to the line joining the points $( 1 , - 2,3 )$ and $( 1,1,0 )$ lies on the plane
(1) $2 x + y - z = 1$
(2) $x - y - 2 z = 1$
(3) $x - 2 y + z = 1$
(4) $x + 2 y - z = 1$

If $( a , b , c )$ is the image of the point $( 1,2 , - 3 )$ in the line, $\frac { x + 1 } { 2 } = \frac { y - 3 } { - 2 } = \frac { z } { - 1 }$, then $a + b + c$ is equal to:
(1) 2
(2) - 1
(3) 3
(4) 1

Consider the line $L$ given by the equation $\frac { x - 3 } { 2 } = \frac { y - 1 } { 1 } = \frac { z - 2 } { 1 }$. Let $Q$ be the mirror image of the point $( 2,3 , - 1 )$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$, and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?
(1) $( - 1,1,2 )$
(2) $( 1,1,1 )$
(3) $( 1,1,2 )$
(4) $( 1,2,2 )$

Let the foot of perpendicular of the point $P ( 3 , - 2 , - 9 )$ on the plane passing through the points $( - 1 , - 2 , - 3 ) , ( 9,3,4 ) , ( 9 , - 2,1 )$ be $Q ( \alpha , \beta , \gamma )$. Then the distance of $Q$ from the origin is
(1) $\sqrt { 42 }$
(2) $\sqrt { 38 }$
(3) $\sqrt { 35 }$
(4) $\sqrt { 29 }$

The foot of perpendicular of the point $( 2,0,5 )$ on the line $\frac { x + 1 } { 2 } = \frac { y - 1 } { 5 } = \frac { z + 1 } { - 1 }$ is $( \alpha , \beta , \gamma )$. Then, which of the following is NOT correct?
(1) $\frac { \alpha \beta } { \gamma } = \frac { 4 } { 15 }$
(2) $\frac { \alpha } { \beta } = - 8$
(3) $\frac { \beta } { \gamma } = - 5$
(4) $\frac { \gamma } { \alpha } = \frac { 5 } { 8 }$

Let the plane $x + 3 y - 2 z + 6 = 0$ meet the co-ordinate axes at the points $A , B , C$. If the orthocenter of the triangle $A B C$ is $\left( \alpha , \beta , \frac { 6 } { 7 } \right)$, then $98 ( \alpha + \beta ) ^ { 2 }$ is equal to $\_\_\_\_$ .

If the mirror image of the point $P(3,4,9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then $14\alpha + \beta + \gamma$ is:
(1) 102
(2) 138
(3) 108
(4) 132

Let the image of the point $( 1,0,7 )$ in the line $\frac { x } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 2 } { 3 }$ be the point $( \alpha , \beta , \gamma )$. Then which one of the following points lies on the line passing through $( \alpha , \beta , \gamma )$ and making angles $\frac { 2 \pi } { 3 }$ and $\frac { 3 \pi } { 4 }$ with y - axis and z axis respectively and an acute angle with x -axis?
(1) $( 1 , - 2,1 + \sqrt { 2 } )$
(2) $( 1,2,1 - \sqrt { 2 } )$
(3) $( 3,4,3 - 2 \sqrt { 2 } )$
(4) $( 3 , - 4,3 + 2 \sqrt { 2 } )$

Consider a line L passing through the points $\mathrm { P } ( 1,2,1 )$ and $\mathrm { Q } ( 2,1 , - 1 )$. If the mirror image of the point $\mathrm { A } ( 2,2,2 )$ in the line L is $( \alpha , \beta , \gamma )$, then $\alpha + \beta + 6 \gamma$ is equal to $\_\_\_\_$

Let $( \alpha , \beta , \gamma )$ be the image of the point $( 8,5,7 )$ in the line $\frac { x - 1 } { 2 } = \frac { y + 1 } { 3 } = \frac { z - 2 } { 5 }$. Then $\alpha + \beta + \gamma$ is equal to :
(1) 16
(2) 20
(3) 14
(4) 18

Let $\mathrm { P } ( \alpha , \beta , \gamma )$ be the image of the point $\mathrm { Q } ( 1,6,4 )$ in the line $\frac { x } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 2 } { 3 }$. Then $2 \alpha + \beta + \gamma$ is equal to $\_\_\_\_$

Let P be the image of the point $\mathrm{Q}(7, -2, 5)$ in the line $\mathrm{L} : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $\mathrm{R}(5, \mathrm{p}, \mathrm{q})$ be a point on $L$. Then the square of the area of $\triangle PQR$ is $\_\_\_\_$.

Let $E: x + z = 2$ be the plane in coordinate space passing through the three points $A(2,-1,0)$, $B(0,1,2)$, $C(-2,1,4)$. There is another point $P$ on the plane $z = 1$ whose projection onto $E$ is equidistant from points $A$, $B$, and $C$. Then the distance from point $P$ to plane $E$ is (16--1)$\sqrt{16\text{-}2}$. (Express as a simplified radical)