The lines $\frac { x - 2 } { 2 } = \frac { y } { - 2 } = \frac { z - 7 } { 16 }$ and $\frac { x + 3 } { 4 } = \frac { y + 2 } { 3 } = \frac { z + 2 } { 1 }$ intersect at the point $P$. If the distance of $P$ from the line $\frac { \mathrm { x } + 1 } { 2 } = \frac { \mathrm { y } - 1 } { 3 } = \frac { \mathrm { z } - 1 } { 1 }$ is $l$, then $14 l ^ { 2 }$ is equal to $\_\_\_\_$ .

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 $\_\_\_\_$

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 $\_\_\_\_$

The distance of the line $\frac { x - 2 } { 2 } = \frac { y - 6 } { 3 } = \frac { z - 3 } { 4 }$ from the point $( 1,4,0 )$ along the line $\frac { x } { 1 } = \frac { y - 2 } { 2 } = \frac { z + 3 } { 3 }$ is :
(1) $\sqrt { 17 }$
(2) $\sqrt { 15 }$
(3) $\sqrt { 14 }$
(4) $\sqrt { 13 }$

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 $\_\_\_\_$.

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 tetrahedron $A B C D$, $\overline { A B } = \overline { A C } = \overline { A D } = 4 \sqrt { 6 }$, $\overline { B D } = \overline { C D } = 8$, and $\cos \angle B A C = \frac { 1 } { 3 }$. The distance from point $D$ to plane $A B C$ is (31) $\sqrt { (32) }$. (Express as a fraction in simplest radical form)

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 $\gamma$)

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 $E_{1}$. If a fourth plane $E_{4}$ together with $E_{1}, E_{2}, E_{3}$ encloses a regular tetrahedron with edge length $6\sqrt{2}$, find the equation of $E_{4}$ (write in the form $x + ay + bz = c$).

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) $\frac{\sqrt{3}}{2}$

In coordinate space, let $O$ be the origin and $E$ be the plane $x - z = 4$.
It is known that there is a point $P(a, b, c)$ in space such that the angle $\theta$ between vector $\overrightarrow{OP}$ and vector $(1, 0, 0)$ satisfies $\theta \leq \frac{\pi}{6}$. Show that the real numbers $a, b, c$ satisfy the inequality $a^{2} \geq 3\left(b^{2} + c^{2}\right)$. (Non-multiple choice question, 4 points)

In coordinate space, let $O$ be the origin and $E$ be the plane $x - z = 4$.
Continuing from question 19, it is known that point $P$ is on plane $E$ and $b = 0$. Find the maximum possible range of $c$ and the minimum possible length of line segment $\overline{OP}$. (Non-multiple choice question, 8 points)

In space, there is a regular cube $ABCD-EFGH$, where vertices $A, B, C, D$ lie on the same plane, and $\overline{AE}$ is one of its edges, as shown in the figure. Among the following options, select the plane that is perpendicular to both plane $BGH$ and plane $CFE$.
(1) Plane $ADH$
(2) Plane $BCD$
(3) Plane $CDG$
(4) Plane $DFG$
(5) Plane $DFH$

In coordinate space, point $A$ has coordinates $(a, b, c)$, where $a, b, c$ are all negative real numbers. Point $A$ is at distance 6 from each of the three planes $E _ { 1 } : 4 y + 3 z = 2$, $E _ { 2 } : 3 y + 4 z = - 5$, and $E _ { 3 } : x + 2 y + 2 z = - 2$. Then $a + b + c =$ (14-1) (14-2) (14-3).