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 $\_\_\_\_$

Q78. If the shortest distance between the lines $\begin{aligned} & L _ { 1 } : \vec { r } = ( 2 + \lambda ) \hat { i } + ( 1 - 3 \lambda ) \hat { j } + ( 3 + 4 \lambda ) \hat { k } , \quad \lambda \in \mathbb { R } \\ & L _ { 2 } : \vec { r } = 2 ( 1 + \mu ) \hat { i } + 3 ( 1 + \mu ) \hat { j } + ( 5 + \mu ) \hat { k } , \quad \mu \in \mathbb { R } \end{aligned}$ is $\frac { m } { \sqrt { n } }$ , where $\operatorname { gcd } ( m , n ) = 1$, then the value of $m + n$ equals
(1) 390
(2) 384
(3) 377
(4) 387

Q79. 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

Q79. If the shortest distance between the lines $\frac { x - \lambda } { 2 } = \frac { y - 4 } { 3 } = \frac { z - 3 } { 4 }$ and $\frac { x - 2 } { 4 } = \frac { y - 4 } { 6 } = \frac { z - 7 } { 8 }$ is $\frac { 13 } { \sqrt { 29 } }$, then a value of $\lambda$ is :
(1) - 1
(2) $- \frac { 13 } { 25 }$
(3) $\frac { 13 } { 25 }$
(4) 1

Q79. Let the line L intersect the lines $x - 2 = - y = z - 1,2 ( x + 1 ) = 2 ( y - 1 ) = z + 1$ and be parallel to the line $\frac { x - 2 } { 3 } = \frac { y - 1 } { 1 } = \frac { z - 2 } { 2 }$. Then which of the following points lies on L ?
(1) $\left( - \frac { 1 } { 3 } , 1 , - 1 \right)$
(2) $\left( - \frac { 1 } { 3 } , - 1,1 \right)$
(3) $\left( - \frac { 1 } { 3 } , 1,1 \right)$
(4) $\left( - \frac { 1 } { 3 } , - 1 , - 1 \right)$

Q80. The shortest distance between the lines $\frac { x - 3 } { 4 } = \frac { y + 7 } { - 11 } = \frac { z - 1 } { 5 }$ and $\frac { x - 5 } { 3 } = \frac { y - 9 } { - 6 } = \frac { z + 2 } { 1 }$ is:
(1) $\frac { 178 } { \sqrt { 563 } }$
(2) $\frac { 187 } { \sqrt { 563 } }$
(3) $\frac { 185 } { \sqrt { 563 } }$
(4) $\frac { 179 } { \sqrt { 563 } }$

Q89. If the shortest distance between the lines $\frac { x - \lambda } { 3 } = \frac { y - 2 } { - 1 } = \frac { z - 1 } { 1 }$ and $\frac { x + 2 } { - 3 } = \frac { y + 5 } { 2 } = \frac { z - 4 } { 4 }$ is $\frac { 44 } { \sqrt { 30 } }$, then the largest possible value of $| \lambda |$ is equal to $\_\_\_\_$

Q90. If the shortest distance between the lines $\frac { x + 2 } { 2 } = \frac { y + 3 } { 3 } = \frac { z - 5 } { 4 }$ and $\frac { x - 3 } { 1 } = \frac { y - 2 } { - 3 } = \frac { z + 4 } { 2 }$ is $\frac { 38 } { 3 \sqrt { 5 } } \mathrm { k }$, and $\int _ { 0 } ^ { \mathrm { k } } \left[ x ^ { 2 } \right] \mathrm { d } x = \alpha - \sqrt { \alpha }$, where $[ x ]$ denotes the greatest integer function, then $6 \alpha ^ { 3 }$ is equal to $\_\_\_\_$墐

ANSWER KEYS

\begin{tabular}{|l|l|} \hline 1. (3) & 2. (1) \hline 9. (3) & 10. (3) \hline 17. (1) & 18. (2) \hline 25. (16) & 26. (1) \hline 33. (4) & 34. (3) \hline 41. (4) & 42. (1) \hline 49. (4) & 50. (1) \hline 57. (100) & 58. (4) \hline 65. (1) & 66. (3) \hline

Q90. Let the point $( - 1 , \alpha , \beta )$ lie on the line of the shortest distance between the lines $\frac { x + 2 } { - 3 } = \frac { y - 2 } { 4 } = \frac { z - 5 } { 2 }$ and $\frac { x + 2 } { - 1 } = \frac { y + 6 } { 2 } = \frac { z - 1 } { 0 }$. Then $( \alpha - \beta ) ^ { 2 }$ is equal to $\_\_\_\_$

ANSWER KEYS

\begin{tabular}{|l|l|l|l|} \hline 1. (3) & 2. (3) & 3. (4) & 4. (2) \hline 9. (3) & 10. (4) & 11. (3) & 12. (2) \hline 17. (2) & 18. (2) & 19. (3) & 20. (2) \hline 25. (16) & 26. (5) & 27. (500) & 28. (4) \hline

Q90. 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 $\_\_\_\_$

ANSWER KEYS

\begin{tabular}{|l|l|l|l|} \hline 1. (4) & 2. (1) & 3. (3) & 4. (3) \hline 9. (3) & 10. (1) & 11. (4) & 12. (1) \hline 17. (2) & 18. (2) & 19. (3) & 20. (3) \hline 25. (5) & 26. (5) & 27. (10) & 28. (22) \hline 33. (2) & 34. (2) & 35. (3) & 36. (3) \hline 41. (3) & 42. (2) & 43. (3) & 44. (4) \hline

Given the line $r \equiv \frac { x - 1 } { 2 } = \frac { y - 3 } { - 2 } = z$ and the line s that passes through the point $( 2 ; - 5 ; 1 )$ and has direction $( - 1 ; 0 ; - 1 )$, find: a) (1 point) Study the relative position of the two lines. b) (1 point) Calculate a plane that is parallel to r and contains s. c) ( 0.5 points) Calculate a plane perpendicular to the line r and that passes through the origin of coordinates.

Given the points $\mathrm { P } ( - 3,1,2 )$ and $\mathrm { Q } ( - 1,0,1 )$ and the plane $\pi$ with equation $x + 2 y - 3 z = 4$, it is requested:\ a) (1 point) Find the projection of $Q$ onto $\pi$.\ b) (0.5 points) Write the equation of the plane parallel to $\pi$ that passes through point $P$.\ c) (1 point) Write the equation of the plane perpendicular to $\pi$ that contains points $P$ and $Q$.

Given the point $A ( 1,0 , - 1 )$, the line $r \equiv x - 1 = y + 1 = \frac { z - 2 } { 2 }$ and the plane $\pi \equiv x + y - z = 6$, find:
a) ( 0.75 points) Find the angle formed by the plane $\pi$ and the plane perpendicular to the line $r$ that passes through point $A$.
b) ( 0.75 points) Determine the distance between the line $r$ and the plane $\pi$.
c) (1 point) Calculate an equation of the line that passes through A, forms a right angle with the line $r$ and does not intersect the plane $\pi$.

Given the lines
$$r \equiv \frac { x - 2 } { 1 } = \frac { y + 1 } { 1 } = \frac { z + 4 } { - 3 } , \quad s \equiv \left\{ \begin{array} { l } x + z = 2 \\ - 2 x + y - 2 z = 1 \end{array} . \right.$$
a) (1.5 points) Write an equation of the common perpendicular line to $r$ and $s$.
b) (1 point) Calculate the distance between $r$ and $s$.

Let the line $r \equiv \left\{ \begin{array} { l } - x - y + z = 0 \\ 2 x + 3 y - z + 1 = 0 \end{array} \right.$ and the plane $\pi \equiv 2 x + y - z + 3 = 0$. It is requested:\ a) ( 0.75 points) Calculate the angle formed by $r$ and $\pi$.\ b) (1 point) Find the symmetric point of the intersection of line $r$ and plane $\pi$ with respect to the plane $z - y = 0$.\ c) ( 0.75 points) Determine the orthogonal projection of line $r$ onto plane $\pi$.

Given the line $r \equiv \frac { x - 1 } { 2 } = \frac { y } { 1 } = \frac { z + 1 } { - 2 }$, the plane $\pi : x - z = 2$ and the point A(1,1,1), find:\ a) ( 0.75 points) Study the relative position of $r$ and $\pi$ and calculate their intersection, if it exists.\ b) ( 0.75 points) Calculate the orthogonal projection of point A onto the plane $\pi$.\ c) (1 point) Calculate the point symmetric to point A with respect to the line $r$.

Let the line $r \equiv \left\{ \begin{array} { l } x = \lambda \\ y = 0 \\ z = 0 \end{array} \right.$ and the plane $\pi : z = 0$.
a) (1 point) Find an equation of the line parallel to the plane $\pi$ whose direction is perpendicular to $r$ and passes through the point $( 1,1,1 )$.
b) (1.5 points) Find an equation of a line that forms an angle of $\frac { \pi } { 4 }$ radians with the line $r$, is contained in the plane $\pi$ and passes through the point ( $0,0,0$ ).

11. Let the equations of three lines $L _ { 1 } , L _ { 2 } , L _ { 3 }$ in coordinate space be
$$L _ { 1 } : \frac { x } { 1 } = \frac { y + 3 } { 6 } = \frac { z + 4 } { 8 } ; \quad L _ { 2 } : \frac { x } { 1 } = \frac { y + 3 } { 3 } = \frac { z + 4 } { 4 } ; \quad L _ { 3 } : \frac { x } { 1 } = \frac { y } { 3 } = \frac { z } { 4 }$$
Which of the following statements are correct?
(1) $L _ { 1 }$ and $L _ { 2 }$ intersect
(2) $L _ { 2 }$ and $L _ { 3 }$ are parallel
(3) The distance between points $P ( 0 , - 3 , - 4 )$ and $Q ( 0,0,0 )$ equals the shortest distance from point $P$ to $L _ { 3 }$
(4) The line $L$ : $\left\{ \begin{array} { c } x = 0 \\ \frac { y + 3 } { 4 } = \frac { z + 4 } { - 3 } \end{array} \right.$ is perpendicular to both $L _ { 1 }$ and $L _ { 2 }$
(5) The three lines $L _ { 1 } , L _ { 2 } , L _ { 3 }$ are coplanar

In coordinate space, let $E$ be the plane passing through the three points $A ( 0 , - 1 , - 1 )$ , $B ( 1 , - 1 , - 2 )$ , $C ( 0,1,0 )$ . Assume $H$ is a point in space satisfying $\overrightarrow { AH } = \frac { 2 } { 3 } \overrightarrow { AB } - \frac { 1 } { 3 } \overrightarrow { AC } + 3 ( \overrightarrow { AB } \times \overrightarrow { AC } )$ . Based on the above, answer the following questions.
(1) Find the volume of tetrahedron $ABCH$ . (4 points) (Note: The volume of a tetrahedron is one-third of the base area times the height)
(2) Let $H ^ { \prime }$ be the symmetric point of point $H$ with respect to plane $E$ . Find the coordinates of $H ^ { \prime }$ . (4 points)
(3) Determine whether the projection of point $H ^ { \prime }$ onto plane $E$ lies inside $\triangle ABC$ . Explain your reasoning. (4 points) (Note: The interior of a triangle does not include the three sides of the triangle)

In coordinate space, there are two lines $L _ { 1 } , L _ { 2 }$ and a plane $E$. The line $L _ { 1 } : \frac { x } { 2 } = \frac { y } { - 3 } = \frac { z } { - 5 }$, and the parametric equation of $L _ { 2 }$ is $\left\{ \begin{array} { l } x = 1 \\ y = 1 + 2 t \\ z = 1 + 3 t \end{array} \right.$ ($t$ is a real number). If $L _ { 1 }$ lies on $E$ and $L _ { 2 }$ does not intersect $E$, then the equation of $E$ is $x -$ (16) $y +$ (17) $z =$ (18).

In coordinate space, on the plane $x - y + 2 z = 3$ there are two distinct lines $L : \frac { x } { 2 } - 1 = y + 1 = - 2 z$ and $L ^ { \prime }$ . It is known that $L$ also lies on another plane $E$ , and the projection of $L ^ { \prime }$ on $E$ coincides with $L$ . Then the equation of $E$ is $x +$ (16-1)(16-2) $y +$ (16-3)(16-4) $z =$ (16-5) .

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)

In coordinate space, there are two non-intersecting lines $L_{1}: \left\{\begin{array}{l} x = 1+t \\ y = 1-t \\ z = 2+t \end{array}\right.$, $t$ is a real number, $L_{2}: \left\{\begin{array}{l} x = 2+2s \\ y = 5+s \\ z = 6-s \end{array}\right.$, $s$ is a real number. Another line $L_{3}$ intersects both $L_{1}$ and $L_{2}$ and is perpendicular to both. If points $P$ and $Q$ are on $L_{1}$ and $L_{2}$ respectively and both are at distance 3 from $L_{3}$, then the distance between points $P$ and $Q$ is (17--1)$\sqrt{17\text{-}2}$. (Express as a simplified radical)

In coordinate space, consider three planes $E_{1}: x + y + z = 7$, $E_{2}: x - y + z = 3$, $E_{3}: x - y - z = -5$. Let $L_{3}$ be the line of intersection of $E_{1}$ and $E_{2}$; $L_{1}$ be the line of intersection of $E_{2}$ and $E_{3}$; $L_{2}$ be the line of intersection of $E_{3}$ and $E_{1}$. It is known that the three lines $L_{1}, L_{2}, L_{3}$ have a common intersection point. Find the coordinates of this common intersection point $P$.

In coordinate space, consider three planes $E_{1}: x + y + z = 7$, $E_{2}: x - y + z = 3$, $E_{3}: x - y - z = -5$. Let $L_{3}$ be the line of intersection of $E_{1}$ and $E_{2}$; $L_{1}$ be the line of intersection of $E_{2}$ and $E_{3}$; $L_{2}$ be the line of intersection of $E_{3}$ and $E_{1}$. Explain that among $L_{1}, L_{2}, L_{3}$, the acute angle between any two lines is $60^{\circ}$. (Note: Let the acute angle between $L_{1}$ and $L_{2}$ be $\alpha$, the acute angle between $L_{2}$ and $L_{3}$ be $\beta$, and the acute angle between $L_{3}$ and $L_{1}$ be $\gamma$)