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 $\_\_\_\_$.

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

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

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, let $O$ be the origin and $E$ be the plane $x - z = 4$.
If the projection of the origin $O$ onto plane $E$ is point $Q$, and the angle between vector $\overrightarrow{OQ}$ and vector $(1, 0, 0)$ is $\alpha$, what is the value of $\cos\alpha$? (Single choice question, 3 points)
(1) $-\frac{\sqrt{2}}{2}$
(2) $-\frac{1}{2}$
(3) $\frac{1}{2}$
(4) $\frac{\sqrt{2}}{2}$
(5) $\frac{\sqrt{3}}{2}$