Line and Plane Questions in English

(A) Let the point of intersection be $(k, k, k)$.
Since this point lies on the planes $x \sin A + y \sin B + z \sin C = 18$ and $x \sin 2A + y \sin 2B + z \sin 2C = 9$,we have:
$k(\sin A + \sin B + \sin C) = 18 \implies \sin A + \sin B + \sin C = \frac{18}{k}$
$k(\sin 2A + \sin 2B + \sin 2C) = 9 \implies \sin 2A + \sin 2B + \sin 2C = \frac{9}{k}$
Dividing the two equations,we get $\frac{\sin A + \sin B + \sin C}{\sin 2A + \sin 2B + \sin 2C} = \frac{18}{9} = 2$.
Thus,$\sin A + \sin B + \sin C = 2(\sin 2A + \sin 2B + \sin 2C)$.
Using the identities $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ and $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C = 4(2 \sin \frac{A}{2} \cos \frac{A}{2})(2 \sin \frac{B}{2} \cos \frac{B}{2})(2 \sin \frac{C}{2} \cos \frac{C}{2}) = 32 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.
Substituting these into the equation: $4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = 2(32 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2})$.
Canceling $4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ (which is non-zero),we get $1 = 16 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Therefore,$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}$.
Finally,$80 \left( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \right) = 80 \times \frac{1}{16} = 5$.

(C) Let the vertices of the triangle be $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. The midpoint $M$ of $PQ$ is given by $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}) = (2, 1, 2)$.
The centroid $G$ of $\triangle PQR$ is $(\frac{x_1+x_2-1}{3}, \frac{y_1+y_2+4}{3}, \frac{z_1+z_2+2}{3})$.
Substituting the values from $M$,we get $G = (\frac{4-1}{3}, \frac{2+4}{3}, \frac{4+2}{3}) = (1, 2, 2)$.
To find the intersection point $A$ of the lines,let $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1} = k_1$. Then $x = 2, y = 2k_1, z = -k_1-3$.
Let $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1} = k_2$. Then $x = k_2+1, y = -3k_2-3, z = k_2-1$.
Equating $x$: $2 = k_2+1 \implies k_2 = 1$.
Then $y = -3(1)-3 = -6$ and $z = 1-1 = 0$.
Checking in the first line: $y = 2k_1 = -6 \implies k_1 = -3$. Then $z = -(-3)-3 = 0$. The point $A$ is $(2, -6, 0)$.
The distance $AG = \sqrt{(2-1)^2 + (-6-2)^2 + (0-2)^2} = \sqrt{1^2 + (-8)^2 + (-2)^2} = \sqrt{1 + 64 + 4} = \sqrt{69}$.

(D) The line $L_1$ is given by $\frac{x}{1} = \frac{y}{1} = \frac{z-1}{0} = r$. Thus,$Q = (r, r, 1)$.
Since $PQ \perp L_1$,the vector $\vec{PQ} = (r-a, r-a, 1-a)$ is perpendicular to the direction vector $(1, 1, 0)$.
$(r-a)(1) + (r-a)(1) + (1-a)(0) = 0 \implies 2(r-a) = 0 \implies r = a$. So,$Q = (a, a, 1)$.
The line $L_2$ is given by $\frac{x}{1} = \frac{y}{-1} = \frac{z+1}{0} = k$. Thus,$R = (k, -k, -1)$.
Since $PR \perp L_2$,the vector $\vec{PR} = (k-a, -k-a, -1-a)$ is perpendicular to the direction vector $(1, -1, 0)$.
$(k-a)(1) + (-k-a)(-1) + (-1-a)(0) = 0 \implies k-a + k+a = 0 \implies 2k = 0 \implies k = 0$. So,$R = (0, 0, -1)$.
Given $\angle QPR = 90^{\circ}$,we have $\vec{PQ} \cdot \vec{PR} = 0$.
$\vec{PQ} = (a-a, a-a, 1-a) = (0, 0, 1-a)$.
$\vec{PR} = (0-a, 0-a, -1-a) = (-a, -a, -1-a)$.
$(0)(-a) + (0)(-a) + (1-a)(-1-a) = 0$.
$-(1-a)(1+a) = 0 \implies -(1-a^2) = 0 \implies a^2 = 1$.
Therefore,$12a^2 = 12(1) = 12$.

(D) The normal vectors to the planes are $\vec{n_1} = 3\hat{i} - 6\hat{j} - 2\hat{k}$ and $\vec{n_2} = 2\hat{i} + \hat{j} - 2\hat{k}$.
The direction vector $\vec{v}$ of the line of intersection is given by $\vec{n_1} \times \vec{n_2}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & -2 \\ 2 & 1 & -2 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(-6 - (-4)) + \hat{k}(3 - (-12)) = 14\hat{i} + 2\hat{j} + 15\hat{k}$.
Thus,$Statement-2$ is True.
To find a point on the line,set $z = 0$ in the plane equations:
$3x - 6y = 15 \Rightarrow x - 2y = 5$
$2x + y = 5$
Solving these,we get $x = 3, y = -1$. So,$(3, -1, 0)$ is a point on the line.
The equation of the line is $\frac{x-3}{14} = \frac{y+1}{2} = \frac{z-0}{15} = t$.
This gives $x = 14t + 3, y = 2t - 1, z = 15t$.
Comparing this with $Statement-1$,the equations provided are $x = 3 + 14t, y = 1 + 2t, z = 15t$,which is incorrect because the $y$-coordinate is $2t - 1$,not $2t + 1$.
Therefore,$Statement-1$ is False and $Statement-2$ is True.

(B, D, C) $1.$ The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The vector perpendicular to both is $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.
The magnitude is $|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.
The unit vector is $\frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$. Thus,option $(B)$ is correct.
$2.$ The shortest distance between lines passing through $A(-1, -2, -1)$ and $B(2, -2, 3)$ with directions $\vec{v_1}, \vec{v_2}$ is $d = \frac{|(\vec{AB}) \cdot (\vec{v_1} \times \vec{v_2})|}{|\vec{v_1} \times \vec{v_2}|}$.
$\vec{AB} = (2 - (-1))\hat{i} + (-2 - (-2))\hat{j} + (3 - (-1))\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.
$d = \frac{|(3\hat{i} + 0\hat{j} + 4\hat{k}) \cdot (-\hat{i} - 7\hat{j} + 5\hat{k})|}{5\sqrt{3}} = \frac{|-3 + 0 + 20|}{5\sqrt{3}} = \frac{17}{5\sqrt{3}}$. Thus,option $(D)$ is correct.
$3.$ The plane passes through $(-1, -2, -1)$ with normal $\vec{n} = -\hat{i} - 7\hat{j} + 5\hat{k}$.
Equation: $-1(x+1) - 7(y+2) + 5(z+1) = 0 \Rightarrow -x - 1 - 7y - 14 + 5z + 5 = 0 \Rightarrow x + 7y - 5z + 10 = 0$.
The distance from $(1, 1, 1)$ is $d = \frac{|1 + 7(1) - 5(1) + 10|}{\sqrt{1^2 + 7^2 + (-5)^2}} = \frac{|1 + 7 - 5 + 10|}{\sqrt{1 + 49 + 25}} = \frac{13}{\sqrt{75}}$. Thus,option $(C)$ is correct.

(B) The base $OPQR$ is a square in the $xy$-plane with $O(0,0,0)$,$P(3,0,0)$,$Q(3,3,0)$,and $R(0,3,0)$.
The mid-point $T$ of $OQ$ is $(\frac{3+0}{2}, \frac{3+0}{2}, 0) = (1.5, 1.5, 0)$.
Since $S$ is directly above $T$ with $TS=3$,the coordinates of $S$ are $(1.5, 1.5, 3)$.
$(A)$ Vector $\vec{OQ} = 3\hat{i} + 3\hat{j}$ and $\vec{OS} = 1.5\hat{i} + 1.5\hat{j} + 3\hat{k}$.
$|\vec{OQ}| = \sqrt{3^2 + 3^2} = 3\sqrt{2}$. $|\vec{OS}| = \sqrt{1.5^2 + 1.5^2 + 3^2} = \sqrt{2.25 + 2.25 + 9} = \sqrt{13.5} = \sqrt{\frac{27}{2}} = \frac{3\sqrt{3}}{\sqrt{2}}$.
$\vec{OQ} \cdot \vec{OS} = (3)(1.5) + (3)(1.5) + (0)(3) = 4.5 + 4.5 = 9$.
$\cos \theta = \frac{9}{(3\sqrt{2})(\frac{3\sqrt{3}}{\sqrt{2}})} = \frac{9}{9\sqrt{3}} = \frac{1}{\sqrt{3}}$. Thus,$\theta \neq \frac{\pi}{3}$. ($A$ is incorrect).
$(B)$ The plane containing $\triangle OQS$ passes through $O(0,0,0)$,$Q(3,3,0)$,and $S(1.5, 1.5, 3)$.
Normal vector $\vec{n} = \vec{OQ} \times \vec{OS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 0 \\ 1.5 & 1.5 & 3 \end{vmatrix} = \hat{i}(9) - \hat{j}(9) + \hat{k}(0) = 9(\hat{i} - \hat{j})$.
The equation is $1(x-0) - 1(y-0) + 0(z-0) = 0$,i.e.,$x-y=0$. ($B$ is correct).
$(C)$ The perpendicular distance from $P(3,0,0)$ to $x-y=0$ is $d = \frac{|3-0|}{\sqrt{1^2 + (-1)^2}} = \frac{3}{\sqrt{2}}$. ($C$ is correct).
$(D)$ Line $RS$ passes through $R(0,3,0)$ and $S(1.5, 1.5, 3)$. Direction vector $\vec{v} = S-R = (1.5, -1.5, 3) = 1.5(\hat{i} - \hat{j} + 2\hat{k})$.
Line equation: $\vec{r} = (0,3,0) + \lambda(1, -1, 2)$.
Distance from $O(0,0,0)$ to line is $\frac{|\vec{OR} \times \vec{v}|}{|\vec{v}|} = \frac{|(0,3,0) \times (1,-1,2)|}{|(1,-1,2)|} = \frac{|(6, 0, -3)|}{\sqrt{1+1+4}} = \frac{\sqrt{36+9}}{\sqrt{6}} = \sqrt{\frac{45}{6}} = \sqrt{\frac{15}{2}}$. ($D$ is correct).

(C) Let the given point be $Q(3, 1, 7)$. The line passing through $Q$ and perpendicular to the plane $x-y+z=3$ has the direction ratios $(1, -1, 1)$.
Its equation is $\frac{x-3}{1} = \frac{y-1}{-1} = \frac{z-7}{1} = \lambda$.
Any point on this line is $(3+\lambda, 1-\lambda, 7+\lambda)$.
This point lies on the plane $x-y+z=3$,so $(3+\lambda) - (1-\lambda) + (7+\lambda) = 3$.
$3+\lambda-1+\lambda+7+\lambda=3 \Rightarrow 3\lambda+9=3 \Rightarrow 3\lambda=-6 \Rightarrow \lambda=-2$.
The foot of the perpendicular $R$ is $(3-2, 1-(-2), 7-2) = (1, 3, 5)$.
Let $P(x_1, y_1, z_1)$ be the image of $Q$. Since $R$ is the midpoint of $PQ$,we have $\frac{x_1+3}{2}=1, \frac{y_1+1}{2}=3, \frac{z_1+7}{2}=5$.
$x_1 = -1, y_1 = 5, z_1 = 3$. So $P$ is $(-1, 5, 3)$.
The plane passes through $P(-1, 5, 3)$ and contains the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$.
The plane equation is $a(x+1) + b(y-5) + c(z-3) = 0$.
Since it contains the line,it passes through $(0, 0, 0)$,so $a(1) + b(-5) + c(-3) = 0 \Rightarrow a-5b-3c=0$.
Also,the normal vector $(a, b, c)$ is perpendicular to the line direction $(1, 2, 1)$,so $a+2b+c=0$.
Solving $a-5b-3c=0$ and $a+2b+c=0$ by subtraction: $-7b-4c=0 \Rightarrow b = -4c/7$.
Then $a = 5(-4c/7) + 3c = -20c/7 + 21c/7 = c/7$.
Taking $c=7$,we get $a=1, b=-4$. The equation is $1(x+1) - 4(y-5) + 7(z-3) = 0$.
$x+1-4y+20+7z-21 = 0 \Rightarrow x-4y+7z=0$.

(B) Given $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$,so $\triangle UVW$ is an equilateral triangle.
Let $O$ be the origin. The vectors $\vec{u}, \vec{v}, \vec{w}$ are unit vectors,so they lie on a sphere of radius $1$ centered at $O$.
The distance of the plane $P: \sqrt{3}x + 2y + 3z = 16$ from the origin $O(0,0,0)$ is $OQ = \frac{|0+0+0-16|}{\sqrt{(\sqrt{3})^2 + 2^2 + 3^2}} = \frac{16}{\sqrt{3+4+9}} = \frac{16}{4} = 4$.
The distance of any point $(\alpha, \beta, \gamma) \in S$ from the plane $P$ is given as $\frac{7}{2}$. Let $Q$ be the projection of $O$ on $P$. Since $U, V, W$ are at a constant distance from $P$,they lie on a circle which is the intersection of the sphere and a plane parallel to $P$. Let $P'$ be this plane. The distance of $P'$ from $O$ is $OP = OQ - PQ = 4 - \frac{7}{2} = \frac{1}{2}$.
The radius of the circle formed by the intersection of the sphere and plane $P'$ is $R = \sqrt{1^2 - (1/2)^2} = \sqrt{3/4} = \frac{\sqrt{3}}{2}$.
Since $\triangle UVW$ is equilateral and inscribed in this circle of radius $R$,its side length $a = 2R \cos(30^\circ) = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2}$.
The area of $\triangle UVW = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \cdot \frac{9}{4} = \frac{9\sqrt{3}}{16}$.
The volume of the tetrahedron with vertices $O, U, V, W$ is $\frac{1}{3} \times \text{Area}(\triangle UVW) \times OP = \frac{1}{3} \times \frac{9\sqrt{3}}{16} \times \frac{1}{2} = \frac{3\sqrt{3}}{32}$.
The volume $V$ of the parallelepiped determined by $\vec{u}, \vec{v}, \vec{w}$ is $6 \times \text{Volume of tetrahedron} = 6 \times \frac{3\sqrt{3}}{32} = \frac{9\sqrt{3}}{16}$.
Therefore,$\frac{80}{\sqrt{3}} V = \frac{80}{\sqrt{3}} \times \frac{9\sqrt{3}}{16} = 5 \times 9 = 45$.

(B) Given lines are $L_1: \vec{r}_1 = \lambda(\hat{i}+\hat{j}+\hat{k})$ and $L_2: \vec{r}_2 = (\hat{j}-\hat{k}) + \mu(\hat{i}+\hat{k})$.
Any plane $H$ containing $L_1$ passes through the origin $(0,0,0)$ and has a normal vector $\vec{n}$ perpendicular to the direction vector $(1,1,1)$ of $L_1$. Let the plane be $ax+by+cz=0$,where $a+b+c=0$.
The distance $d(H)$ from $L_2$ to $H$ is non-zero only if $L_2$ is parallel to $H$. If $L_2$ is not parallel to $H$,the distance is $0$. To maximize $d(H)$,$L_2$ must be parallel to $H$. Thus,the normal $\vec{n} = (a,b,c)$ must be perpendicular to the direction vector of $L_2$,which is $(1,0,1)$.
So,$a+c=0$. Since $a+b+c=0$ and $a+c=0$,we get $b=0$. Let $a=1$,then $c=-1$. The plane $H_0$ is $x-z=0$.
$(P)$ $d(H_0)$ is the distance from any point on $L_2$ (e.g.,$(0,1,-1)$) to $H_0$: $d = \frac{|0 - (-1)|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$. Thus,$P \rightarrow 5$.
$(Q)$ Distance of $(0,1,2)$ from $x-z=0$ is $\frac{|0-2|}{\sqrt{1^2+(-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Thus,$Q \rightarrow 4$.
$(R)$ Since $H_0$ is $x-z=0$,it passes through the origin $(0,0,0)$. Thus,the distance is $0$. $R \rightarrow 3$.
$(S)$ Intersection of $y=z, x=1, x-z=0$: From $x=1$ and $x-z=0$,$z=1$. From $y=z$,$y=1$. Point is $(1,1,1)$. Distance from origin is $\sqrt{1^2+1^2+1^2} = \sqrt{3}$. Thus,$S \rightarrow 1$.
Therefore,the correct matching is $(P) \rightarrow (5), (Q) \rightarrow (4), (R) \rightarrow (3), (S) \rightarrow (1)$.

(C) The direction vectors of the two lines are $\vec{v_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{v_2} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
The normal vector $\vec{n}$ to the plane containing these lines is $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.
Since the plane passes through the point $(1, 2, 3)$,its equation is $-1(x-1) + 2(y-2) - 1(z-3) = 0$,which simplifies to $-x + 2y - z = 0$ or $x - 2y + z = 0$.
The given plane is $Ax - 2y + z = d$. Comparing the coefficients,we find $A = 1$.
The distance between the parallel planes $x - 2y + z = 0$ and $x - 2y + z = d$ is given by $\frac{|d - 0|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \sqrt{6}$.
$\frac{|d|}{\sqrt{6}} = \sqrt{6} \implies |d| = 6$.

(A) The distance of point $P(x_1, y_1, z_1) = (1, -2, 1)$ from the plane $Ax + By + Cz - D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}$.
Given $d = 5$ and the plane $x + 2y - 2z - \alpha = 0$, we have $5 = \frac{|1(1) + 2(-2) - 2(1) - \alpha|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|1 - 4 - 2 - \alpha|}{\sqrt{9}} = \frac{|-5 - \alpha|}{3}$.
Since $\alpha > 0$, $|-5 - \alpha| = |5 + \alpha| = 5 + \alpha$. Thus, $5 = \frac{5 + \alpha}{3} \Rightarrow 15 = 5 + \alpha \Rightarrow \alpha = 10$.
The equation of the plane is $x + 2y - 2z = 10$.
The line passing through $P(1, -2, 1)$ and perpendicular to the plane has direction ratios $(1, 2, -2)$. Its equation is $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 1}{-2} = k$.
Any point on this line is $(k + 1, 2k - 2, -2k + 1)$.
Since this point lies on the plane, $(k + 1) + 2(2k - 2) - 2(-2k + 1) = 10$.
$k + 1 + 4k - 4 + 4k - 2 = 10 \Rightarrow 9k - 5 = 10 \Rightarrow 9k = 15 \Rightarrow k = \frac{5}{3}$.
Substituting $k = \frac{5}{3}$ into the point coordinates: $x = \frac{5}{3} + 1 = \frac{8}{3}$, $y = 2(\frac{5}{3}) - 2 = \frac{4}{3}$, $z = -2(\frac{5}{3}) + 1 = -\frac{7}{3}$.
The foot of the perpendicular is $\left(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3}\right)$.

(A) The equation of the line $QR$ passing through $Q(2, 3, 5)$ and $R(1, -1, 4)$ is given by $\frac{x-2}{1-2} = \frac{y-3}{-1-3} = \frac{z-5}{4-5}$,which simplifies to $\frac{x-2}{-1} = \frac{y-3}{-4} = \frac{z-5}{-1} = \lambda$.
Thus,any point on the line is $P(2-\lambda, 3-4\lambda, 5-\lambda)$.
Since $P$ lies on the plane $5x - 4y - z = 1$,we have $5(2-\lambda) - 4(3-4\lambda) - (5-\lambda) = 1$.
$10 - 5\lambda - 12 + 16\lambda - 5 + \lambda = 1 \Rightarrow 12\lambda - 7 = 1 \Rightarrow 12\lambda = 8 \Rightarrow \lambda = \frac{2}{3}$.
Substituting $\lambda = \frac{2}{3}$,we get $P = (2-\frac{2}{3}, 3-\frac{8}{3}, 5-\frac{2}{3}) = (\frac{4}{3}, \frac{1}{3}, \frac{13}{3})$.
Now,for the foot of the perpendicular $S$ from $T(2, 1, 4)$ to $QR$,let $S = (2-\mu, 3-4\mu, 5-\mu)$.
The vector $\vec{TS} = (2-\mu-2, 3-4\mu-1, 5-\mu-4) = (-\mu, 2-4\mu, 1-\mu)$.
Since $\vec{TS}$ is perpendicular to the line direction vector $\vec{v} = (-1, -4, -1)$,their dot product is zero: $(-\mu)(-1) + (2-4\mu)(-4) + (1-\mu)(-1) = 0$.
$\mu - 8 + 16\mu - 1 + \mu = 0 \Rightarrow 18\mu = 9 \Rightarrow \mu = \frac{1}{2}$.
Thus,$S = (2-\frac{1}{2}, 3-2, 5-\frac{1}{2}) = (\frac{3}{2}, 1, \frac{9}{2})$.
The distance $PS = \sqrt{(\frac{4}{3}-\frac{3}{2})^2 + (\frac{1}{3}-1)^2 + (\frac{13}{3}-\frac{9}{2})^2} = \sqrt{(-\frac{1}{6})^2 + (-\frac{2}{3})^2 + (-\frac{1}{6})^2} = \sqrt{\frac{1}{36} + \frac{4}{9} + \frac{1}{36}} = \sqrt{\frac{2}{36} + \frac{16}{36}} = \sqrt{\frac{18}{36}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

(A) Let the direction ratios of the line of intersection be $a, b, c$.
Since the line lies in both planes,it is perpendicular to the normals of both planes,$\vec{n_1} = (2, 1, -1)$ and $\vec{n_2} = (1, 2, 1)$.
Thus,the direction vector is $\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1 + 2) - \hat{j}(2 + 1) + \hat{k}(4 - 1) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.
Dividing by $3$,the direction ratios are $1, -1, 1$. Thus,$(A)$ is incorrect.
For $(B)$,the line is $\frac{x - 4/3}{3} = \frac{y - 1/3}{-3} = \frac{z}{3}$,which has direction ratios $3, -3, 3$ or $1, -1, 1$. Since this is parallel to the line of intersection,it is not perpendicular. Thus,$(B)$ is incorrect.
For $(C)$,$\cos \theta = \frac{|(2)(1) + (1)(2) + (-1)(1)|}{\sqrt{2^2+1^2+(-1)^2} \sqrt{1^2+2^2+1^2}} = \frac{|2+2-1|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$. Thus,$\theta = 60^{\circ}$. $(C)$ is correct.
For $(D)$,the normal to $P_3$ is $(1, -1, 1)$. The equation is $1(x-4) - 1(y-2) + 1(z+2) = 0 \Rightarrow x - y + z = 0$.
The distance of $(2, 1, 1)$ from $x - y + z = 0$ is $\frac{|2 - 1 + 1|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{2}{\sqrt{3}}$. $(D)$ is correct.

(A) Let $P \equiv (x_0, y_0, z_0)$.
The line passing through $P$ perpendicular to the plane $x+y=3$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \frac{z-z_0}{0} = k$.
The image $Q$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \frac{z-z_0}{0} = -2 \frac{x_0+y_0-3}{1^2+1^2} = -(x_0+y_0-3)$.
Thus,$x_Q = x_0 - (x_0+y_0-3) = 3-y_0$ and $y_Q = y_0 - (x_0+y_0-3) = 3-x_0$.
Since $Q$ lies on the $z$-axis,$x_Q = 0$ and $y_Q = 0$,which implies $x_0 = 3$ and $y_0 = 3$.
The distance of $P(3, 3, z_0)$ from the $x$-axis is $\sqrt{y_0^2 + z_0^2} = 5$.
Substituting $y_0 = 3$,we get $\sqrt{3^2 + z_0^2} = 5$,so $9 + z_0^2 = 25$,which gives $z_0^2 = 16$,so $z_0 = 4$ (since $P$ is in the first octant).
$P$ is $(3, 3, 4)$. The image $R$ of $P$ in the $xy$-plane is $(3, 3, -4)$.
The length of $PR$ is the distance between $(3, 3, 4)$ and $(3, 3, -4)$,which is $|4 - (-4)| = 8$.

(A) The lines are given by $\overrightarrow{r} = \lambda \hat{i}$,$\overrightarrow{r} = \mu(\hat{i} + \hat{j})$,and $\overrightarrow{r} = v(\hat{i} + \hat{j} + \hat{k})$.
Substituting these into the plane equation $x + y + z = 1$:
For line $A$: $\lambda + 0 + 0 = 1 \Rightarrow \lambda = 1$,so $A = (1, 0, 0)$.
For line $B$: $\mu + \mu + 0 = 1 \Rightarrow 2\mu = 1 \Rightarrow \mu = 1/2$,so $B = (1/2, 1/2, 0)$.
For line $C$: $v + v + v = 1 \Rightarrow 3v = 1 \Rightarrow v = 1/3$,so $C = (1/3, 1/3, 1/3)$.
Vectors are $\overrightarrow{AB} = B - A = (-1/2, 1/2, 0)$ and $\overrightarrow{AC} = C - A = (-2/3, 1/3, 1/3)$.
The cross product $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1/2 & 1/2 & 0 \\ -2/3 & 1/3 & 1/3 \end{vmatrix} = \hat{i}(1/6) - \hat{j}(-1/6) + \hat{k}(-1/6 + 1/3) = (1/6, 1/6, 1/6)$.
The area $\Delta = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \sqrt{(1/6)^2 + (1/6)^2 + (1/6)^2} = \frac{1}{2} \sqrt{3/36} = \frac{\sqrt{3}}{12}$.
Then $(6 \Delta)^2 = (6 \times \frac{\sqrt{3}}{12})^2 = (\frac{\sqrt{3}}{2})^2 = 3/4 = 0.75$.

(D) Let $P = (\lambda, 0, 0)$ be a point on $L_1$,$Q = (0, \mu, 1)$ be a point on $L_2$,and $R = (1, 1, v)$ be a point on $L_3$.
Since $P, Q,$ and $R$ are collinear,the vectors $\vec{PQ}$ and $\vec{QR}$ must be proportional.
$\vec{PQ} = (0 - \lambda, \mu - 0, 1 - 0) = (-\lambda, \mu, 1)$.
$\vec{QR} = (1 - 0, 1 - \mu, v - 1) = (1, 1 - \mu, v - 1)$.
For collinearity,$\frac{-\lambda}{1} = \frac{\mu}{1 - \mu} = \frac{1}{v - 1}$.
This implies $\lambda = -\frac{\mu}{1 - \mu} = \frac{\mu}{\mu - 1}$ and $v - 1 = \frac{1 - \mu}{\mu}$,so $v = 1 + \frac{1 - \mu}{\mu} = \frac{1}{\mu}$.
These values of $\lambda$ and $v$ exist for all $\mu \in R$ except $\mu = 0$ and $\mu = 1$.
If $\mu = 0$,$Q = (0, 0, 1) = \hat{k}$. If $\mu = 1$,$Q = (0, 1, 1) = \hat{j} + \hat{k}$.
Thus,$Q$ can be any point on $L_2$ except $\hat{k}$ and $\hat{j} + \hat{k}$.
Checking the given options:
$(1)$ $\hat{k} + \hat{j}$ (corresponds to $\mu = 1$,excluded)
$(2)$ $\hat{k}$ (corresponds to $\mu = 0$,excluded)
$(3)$ $\hat{k} + \frac{1}{2}\hat{j}$ (corresponds to $\mu = 0.5$,valid)
$(4)$ $\hat{k} - \frac{1}{2}\hat{j}$ (corresponds to $\mu = -0.5$,valid)
Therefore,points $3$ and $4$ are valid.

(D) The point of intersection of $L_1$ and $L_2$ is $(1, 0, 1)$.
Since line $L$ passes through $(1, 0, 1)$,we have $\frac{1-\alpha}{l} = \frac{0-1}{m} = \frac{1-\gamma}{-2}$,which implies $\frac{1-\alpha}{l} = -\frac{1}{m} = \frac{1-\gamma}{-2} = k$ (say). Thus,$1-\alpha = kl$,$m = 1/k$,and $1-\gamma = -2k$.
Direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{v_2} = -3\hat{i} - \hat{j} + \hat{k}$.
The unit vectors are $\hat{u_1} = \frac{\vec{v_1}}{\sqrt{11}}$ and $\hat{u_2} = \frac{\vec{v_2}}{\sqrt{11}}$.
The direction of the angle bisector is $\vec{v} = \hat{u_1} + \hat{u_2} = \frac{1}{\sqrt{11}} ((\hat{i} - \hat{j} + 3\hat{k}) + (-3\hat{i} - \hat{j} + \hat{k})) = \frac{1}{\sqrt{11}} (-2\hat{i} - 2\hat{j} + 4\hat{k}) \propto \hat{i} + \hat{j} - 2\hat{k}$.
Comparing with the direction ratios $(l, m, -2)$ of $L$,we get $\frac{l}{1} = \frac{m}{1} = \frac{-2}{-2} = 1$. Thus,$l=1$ and $m=1$.
Then $l+m = 1+1 = 2$,so $(B)$ is true.
Using $k=1$ in the intersection equations: $1-\alpha = 1 \Rightarrow \alpha=0$ and $1-\gamma = -2 \Rightarrow \gamma=3$.
Then $\alpha-\gamma = 0-3 = -3$. Checking the options,we re-evaluate the bisector direction. The acute angle bisector direction is $\vec{v_1} - \vec{v_2} = 4\hat{i} + 2\hat{k} \propto 2\hat{i} + \hat{k}$ or $\vec{v_1} + \vec{v_2} = -2\hat{i} - 2\hat{j} + 4\hat{k} \propto \hat{i} + \hat{j} - 2\hat{k}$.
Given the denominator of $L$ is $-2$,we use $\vec{v} = \hat{i} + \hat{j} - 2\hat{k}$,so $l=1, m=1$. $\alpha=0, \gamma=3$. $\alpha-\gamma = -3$. Wait,if we use the other bisector,$l=-1, m=-1$,then $l+m=-2$. Re-checking the intersection: $1-\alpha = l/(-1) = -l$,$1-\gamma = -2/(-1) = 2$. $\alpha = 1+l, \gamma = -1$. $\alpha-\gamma = 2+l$. With $l=1, \alpha=2, \gamma=-1, \alpha-\gamma=3$. Thus $(A)$ and $(B)$ are true.

(A) The equation of any plane passing through the line of intersection of the planes $x+2y+3z-2=0$ and $x-y+z-3=0$ is given by $(x+2y+3z-2) + \lambda(x-y+z-3) = 0$.
This simplifies to $(1+\lambda)x + (2-\lambda)y + (3+\lambda)z - (2+3\lambda) = 0$.
The distance of this plane from the point $(3,1,-1)$ is given by $\frac{|(1+\lambda)(3) + (2-\lambda)(1) + (3+\lambda)(-1) - (2+3\lambda)|}{\sqrt{(1+\lambda)^2 + (2-\lambda)^2 + (3+\lambda)^2}} = \frac{2}{\sqrt{3}}$.
Simplifying the numerator: $|3+3\lambda + 2-\lambda - 3-\lambda - 2-3\lambda| = |-2\lambda|$.
Simplifying the denominator: $\sqrt{(1+2\lambda+\lambda^2) + (4-4\lambda+\lambda^2) + (9+6\lambda+\lambda^2)} = \sqrt{3\lambda^2+4\lambda+14}$.
Thus,$\frac{|-2\lambda|}{\sqrt{3\lambda^2+4\lambda+14}} = \frac{2}{\sqrt{3}}$.
Squaring both sides: $\frac{4\lambda^2}{3\lambda^2+4\lambda+14} = \frac{4}{3}$.
$3\lambda^2 = 3\lambda^2+4\lambda+14$,which gives $4\lambda = -14$,so $\lambda = -\frac{7}{2}$.
Substituting $\lambda = -\frac{7}{2}$ into the equation: $(1-\frac{7}{2})x + (2+\frac{7}{2})y + (3-\frac{7}{2})z - (2-\frac{21}{2}) = 0$.
$-\frac{5}{2}x + \frac{11}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$.
Multiplying by $-2$,we get $5x-11y+z-17=0$,or $5x-11y+z=17$.

(B) For two lines to be coplanar,the scalar triple product of the vector connecting a point on each line and the direction vectors of the lines must be zero: $[\vec{a}-\vec{c}, \vec{b}, \vec{d}] = 0$.
Given points $\vec{a} = (1, -1, 0)$ and $\vec{c} = (-1, -1, 0)$,the vector $\vec{a}-\vec{c} = (2, 0, 0)$.
The direction vectors are $\vec{b} = 2\hat{i} + k\hat{j} + 2\hat{k}$ and $\vec{d} = 5\hat{i} + 2\hat{j} + k\hat{k}$.
Setting the determinant to zero:
$\left|\begin{array}{lll} 2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{array}\right| = 0 \Rightarrow 2(k^2 - 4) = 0 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2$.
Case $1$: If $k = 2$,the direction vectors are $\vec{b} = (2, 2, 2)$ and $\vec{d} = (5, 2, 2)$.
The normal vector is $\vec{n}_1 = \vec{b} \times \vec{d} = \left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{array}\right| = 0\hat{i} + 6\hat{j} - 6\hat{k}$.
The plane equation is $(\vec{r} - \vec{a}) \cdot \vec{n}_1 = 0 \Rightarrow 0(x-1) + 6(y+1) - 6(z-0) = 0 \Rightarrow y - z = -1$.
Case $2$: If $k = -2$,the direction vectors are $\vec{b} = (2, -2, 2)$ and $\vec{d} = (5, 2, -2)$.
The normal vector is $\vec{n}_2 = \vec{b} \times \vec{d} = \left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 2 \\ 5 & 2 & -2 \end{array}\right| = 0\hat{i} + 14\hat{j} + 14\hat{k}$.
The plane equation is $(\vec{r} - \vec{a}) \cdot \vec{n}_2 = 0 \Rightarrow 0(x-1) + 14(y+1) + 14(z-0) = 0 \Rightarrow y + z = -1$.
Thus,the planes are $y-z=-1$ and $y+z=-1$,which correspond to options $(C)$ and $(B)$.

(D) Any point on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda$ is given by $(2\lambda-2, -\lambda-1, 3\lambda)$.
Let us choose two points on this line:
For $\lambda=0$,point $A = (-2, -1, 0)$.
For $\lambda=1$,point $B = (0, -2, 3)$.
The foot of the perpendicular $(x, y, z)$ from a point $(x_0, y_0, z_0)$ to the plane $ax+by+cz+d=0$ is given by $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = -\frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}$.
For point $A(-2, -1, 0)$ and plane $x+y+z-3=0$:
$\frac{x+2}{1} = \frac{y+1}{1} = \frac{z-0}{1} = -\frac{-2-1+0-3}{1^2+1^2+1^2} = -\frac{-6}{3} = 2$.
So,$x = 0, y = 1, z = 2$. Point $M = (0, 1, 2)$.
For point $B(0, -2, 3)$ and plane $x+y+z-3=0$:
$\frac{x-0}{1} = \frac{y+2}{1} = \frac{z-3}{1} = -\frac{0-2+3-3}{1^2+1^2+1^2} = -\frac{-2}{3} = \frac{2}{3}$.
So,$x = \frac{2}{3}, y = \frac{2}{3}-2 = -\frac{4}{3}, z = \frac{2}{3}+3 = \frac{11}{3}$. Point $N = (\frac{2}{3}, -\frac{4}{3}, \frac{11}{3})$.
The line passing through $M(0, 1, 2)$ and $N(\frac{2}{3}, -\frac{4}{3}, \frac{11}{3})$ has direction ratios proportional to $(\frac{2}{3}-0, -\frac{4}{3}-1, \frac{11}{3}-2) = (\frac{2}{3}, -\frac{7}{3}, \frac{5}{3})$,which is equivalent to $(2, -7, 5)$.
The equation of the line is $\frac{x-0}{2} = \frac{y-1}{-7} = \frac{z-2}{5}$.

(A) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of $P_1$ and $P_2$. Thus,$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 1 & 2 \\ 3 & 5 & -6 \end{vmatrix} = \hat{i}(-6-10) - \hat{j}(-42-6) + \hat{k}(35-3) = -16\hat{i} + 48\hat{j} + 32\hat{k}$.
Dividing by $-16$,we get the normal vector $\vec{n} = \hat{i} - 3\hat{j} - 2\hat{k}$.
To find the intersection point of $L_1$ and $L_2$,let $L_1: (2k_1+1, -k_1, k_1-3)$ and $L_2: (k_2+4, k_2-3, 2k_2-3)$.
Equating coordinates: $2k_1+1 = k_2+4 \Rightarrow 2k_1 - k_2 = 3$ and $-k_1 = k_2-3 \Rightarrow k_1+k_2 = 3$.
Adding these gives $3k_1 = 6 \Rightarrow k_1 = 2$. Then $k_2 = 1$.
Point of intersection: $(2(2)+1, -2, 2-3) = (5, -2, -1)$.
The plane equation is $1(x-5) - 3(y+2) - 2(z+1) = 0 \Rightarrow x - 3y - 2z - 5 - 6 - 2 = 0 \Rightarrow x - 3y - 2z = 13$.
Comparing with $ax+by+cz=d$,we get $a=1, b=-3, c=-2, d=13$.
Thus,$P-3, Q-2, R-4, S-1$. The correct option is $A$.

(C) The first line is $L_1: \frac{x}{1} = \frac{y}{1}, z=1$. Let any point $Q$ on $L_1$ be $(\alpha, \alpha, 1)$.
Direction ratios of $PQ$ are $(\alpha-\lambda, \alpha-\lambda, 1-\lambda)$.
Since $PQ$ is perpendicular to $L_1$ (direction vector $(1, 1, 0)$),we have $(\alpha-\lambda)(1) + (\alpha-\lambda)(1) + (1-\lambda)(0) = 0$,which gives $2(\alpha-\lambda) = 0$,so $\alpha = \lambda$.
Thus,$Q = (\lambda, \lambda, 1)$ and the vector $\vec{PQ} = (0, 0, 1-\lambda)$.
The second line is $L_2: \frac{x}{-1} = \frac{y}{1}, z=-1$. Let any point $R$ on $L_2$ be $(-\beta, \beta, -1)$.
Direction ratios of $PR$ are $(-\beta-\lambda, \beta-\lambda, -1-\lambda)$.
Since $PR$ is perpendicular to $L_2$ (direction vector $(-1, 1, 0)$),we have $(-\beta-\lambda)(-1) + (\beta-\lambda)(1) + (-1-\lambda)(0) = 0$,which gives $\beta+\lambda+\beta-\lambda = 0$,so $2\beta = 0$,hence $\beta = 0$.
Thus,$R = (0, 0, -1)$ and the vector $\vec{PR} = (-\lambda, -\lambda, -1-\lambda)$.
Given $\angle QPR = 90^\circ$,so $\vec{PQ} \cdot \vec{PR} = 0$.
$(0)(-\lambda) + (0)(-\lambda) + (1-\lambda)(-1-\lambda) = 0$.
$-(1-\lambda)(1+\lambda) = 0 \Rightarrow \lambda^2 - 1 = 0 \Rightarrow \lambda = \pm 1$.
If $\lambda = 1$,$P = (1, 1, 1)$,which lies on $L_1$,so $PQ$ is not uniquely defined. Thus,$\lambda = -1$ is the only solution.

(C) The equation of any plane passing through the intersection of $P_1: y=0$ and $P_2: x+z-1=0$ is given by $(x+z-1) + \lambda y = 0$,which simplifies to $x + \lambda y + z - 1 = 0$.
The distance of the point $(0, 1, 0)$ from this plane is given as $1$:
$\frac{|0 + \lambda(1) + 0 - 1|}{\sqrt{1^2 + \lambda^2 + 1^2}} = 1$
$\frac{|\lambda - 1|}{\sqrt{\lambda^2 + 2}} = 1$
Squaring both sides: $(\lambda - 1)^2 = \lambda^2 + 2$
$\lambda^2 - 2\lambda + 1 = \lambda^2 + 2$
$-2\lambda = 1 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the plane equation:
$x - \frac{1}{2}y + z - 1 = 0 \Rightarrow 2x - y + 2z - 2 = 0$.
Now,the distance of the point $(\alpha, \beta, \gamma)$ from the plane $2x - y + 2z - 2 = 0$ is $2$:
$\frac{|2\alpha - \beta + 2\gamma - 2|}{\sqrt{2^2 + (-1)^2 + 2^2}} = 2$
$\frac{|2\alpha - \beta + 2\gamma - 2|}{3} = 2$
$|2\alpha - \beta + 2\gamma - 2| = 6$.
This gives two possibilities:
$2\alpha - \beta + 2\gamma - 2 = 6 \Rightarrow 2\alpha - \beta + 2\gamma - 8 = 0$
$2\alpha - \beta + 2\gamma - 2 = -6 \Rightarrow 2\alpha - \beta + 2\gamma + 4 = 0$.
Thus,options $(B)$ and $(D)$ are correct.

(A) The line $L$ passes through the origin $(0, 0, 0)$ and is equidistant from planes $P_1$ and $P_2$. Thus,$L$ must be parallel to the line of intersection of $P_1$ and $P_2$.
Let the direction ratios of $L$ be $(a, b, c)$. Since $L$ is parallel to the intersection of $P_1$ and $P_2$,its direction vector is parallel to $\vec{n}_1 \times \vec{n}_2$,where $\vec{n}_1 = (1, 2, -1)$ and $\vec{n}_2 = (2, -1, 1)$.
$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(1+2) + \hat{k}(-1-4) = (1, -3, -5)$.
So,the equation of line $L$ is $\frac{x}{1} = \frac{y}{-3} = \frac{z}{-5} = k$.
$M$ is the locus of the feet of the perpendiculars from points on $L$ to plane $P_1$. This is the projection of line $L$ onto plane $P_1$.
The projection of a line onto a plane is a line. The point on $L$ at $k=0$ is $(0, 0, 0)$. The foot of the perpendicular from $(0, 0, 0)$ to $P_1: x+2y-z+1=0$ is given by $\frac{x-0}{1} = \frac{y-0}{2} = \frac{z-0}{-1} = -\frac{0+0-0+1}{1^2+2^2+(-1)^2} = -\frac{1}{6}$.
Thus,the foot is $\left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$,which is point $(B)$.
Since $M$ is a line passing through $(B)$ with direction vector equal to the projection of $(1, -3, -5)$ onto $P_1$,we check the points. Point $(A) \left(0, -\frac{5}{6}, -\frac{2}{3}\right)$ lies on $P_1$ because $0 + 2(-\frac{5}{6}) - (-\frac{2}{3}) + 1 = -\frac{5}{3} + \frac{2}{3} + 1 = 0$. Testing if the vector from $B$ to $A$ is parallel to the projection direction confirms $(A)$ lies on $M$.

(A) The line of intersection of the two planes $P_1$ and $P_2$ is found by solving the system of equations:
$10x + 15y + 12z = 60$
$-2x + 5y + 4z = 20$
Multiplying the second equation by $5$,we get $-10x + 25y + 20z = 100$. Adding this to the first equation gives $40y + 32z = 160$,or $5y + 4z = 20$.
Setting $z = 5k$,we get $5y = 20 - 20k$,so $y = 4 - 4k$. Substituting into the second plane equation: $-2x + 5(4 - 4k) + 4(5k) = 20 \implies -2x + 20 - 20k + 20k = 20 \implies x = 0$.
The line of intersection is $\frac{x}{0} = \frac{y-4}{-4} = \frac{z}{5}$.
An edge of a tetrahedron whose two faces lie on $P_1$ and $P_2$ must either be skew to the line of intersection or intersect it at a point not on the planes (or lie within one of the planes).
Checking the options,lines $A$,$B$,and $C$ satisfy the conditions to be edges of such a tetrahedron.

(D) The equation of the plane is $\vec{r} = \hat{k} + t(-\hat{i} + \hat{j}) + p(-\hat{i} + \hat{k})$.
This represents a plane passing through $(0, 0, 1)$ with normal vector $\vec{n} = (-\hat{i} + \hat{j}) \times (-\hat{i} + \hat{k}) = \hat{i} + \hat{j} + \hat{k}$.
The equation of the plane is $1(x-0) + 1(y-0) + 1(z-1) = 0$,which simplifies to $x + y + z = 1$.
Given $Q = (10, 15, 20)$ and $S = (\alpha, \beta, \gamma)$,the reflection formula is $\frac{\alpha-10}{1} = \frac{\beta-15}{1} = \frac{\gamma-20}{1} = -2 \frac{10+15+20-1}{1^2+1^2+1^2} = -2 \frac{44}{3} = -\frac{88}{3}$.
Thus,$\alpha = 10 - \frac{88}{3} = -\frac{58}{3}$,$\beta = 15 - \frac{88}{3} = -\frac{43}{3}$,and $\gamma = 20 - \frac{88}{3} = -\frac{28}{3}$.
Checking the options:
$(A)$ $3(\alpha+\beta) = 3(-\frac{58}{3} - \frac{43}{3}) = -101$ (True).
$(B)$ $3(\beta+\gamma) = 3(-\frac{43}{3} - \frac{28}{3}) = -71$ (True).
$(C)$ $3(\gamma+\alpha) = 3(-\frac{28}{3} - \frac{58}{3}) = -86$ (True).
$(D)$ $3(\alpha+\beta+\gamma) = 3(-\frac{58+43+28}{3}) = -129$ (False).
Therefore,options $A, B, C$ are correct.

(C) Given lines are $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3} = a$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma} = b$.
For $L_1$,$x = a-11, y = 2a-21, z = 3a-29$.
For $L_2$,$x = 3b-16, y = 2b-11, z = b\gamma-4$.
Equating coordinates for intersection:
$a-11 = 3b-16 \Rightarrow a-3b = -5$ (Eq. $1$)
$2a-21 = 2b-11 \Rightarrow 2a-2b = 10 \Rightarrow a-b = 5$ (Eq. $2$)
Solving $(1)$ and $(2)$: $a=10, b=5$.
Substitute into $z$ coordinates: $3(10)-29 = 5\gamma-4 \Rightarrow 1 = 5\gamma-4 \Rightarrow 5\gamma = 5 \Rightarrow \gamma = 1$.
Intersection point $R_1 = (10-11, 20-21, 30-29) = (-1, -1, 1)$.
Thus,$\vec{OR_1} = -\hat{i}-\hat{j}+\hat{k}$.
Normal vector $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 2 & \gamma \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} = \hat{i}(2-6) - \hat{j}(1-9) + \hat{k}(2-6) = -4\hat{i} + 8\hat{j} - 4\hat{k}$.
Unit normal $\hat{n} = \pm \frac{-4\hat{i} + 8\hat{j} - 4\hat{k}}{\sqrt{16+64+16}} = \pm \frac{-4\hat{i} + 8\hat{j} - 4\hat{k}}{\sqrt{96}} = \pm \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}}$.
Matching with options,we take $\hat{n} = \frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}$.
$\vec{OR_1} \cdot \hat{n} = (-\hat{i}-\hat{j}+\hat{k}) \cdot (\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}) = \frac{-1+2+1}{\sqrt{6}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}}$.
Therefore,$(P) \rightarrow (3), (Q) \rightarrow (4), (R) \rightarrow (1), (S) \rightarrow (5)$.

(B) Let the line $L_1$ be $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4} = k$. Any point on $L_1$ is $(2k+2, 3k+6, 4k+3)$.
Let the line $L_2$ passing through $(1,4,0)$ be $\frac{x-1}{1}=\frac{y-4}{2}=\frac{z-0}{3} = t$. Any point on $L_2$ is $(t+1, 2t+4, 3t)$.
For the point of intersection,we equate the coordinates:
$2k+2 = t+1 \Rightarrow t = 2k+1$
$3k+6 = 2t+4 \Rightarrow 3k+6 = 2(2k+1)+4 = 4k+6 \Rightarrow k=0$.
Substituting $k=0$ into the coordinates of $L_1$,we get the point of intersection $P = (2,6,3)$.
The distance from $(1,4,0)$ to $(2,6,3)$ is $\sqrt{(2-1)^2 + (6-4)^2 + (3-0)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}$.

(A) The line $L_1$ is given by $\overrightarrow{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} + \hat{k}) = (\lambda - 1)\hat{i} + (2\lambda + 2)\hat{j} + (\lambda + 1)\hat{k}$.
The line $L_2$ is given by $\overrightarrow{r} = (\hat{j} + \hat{k}) + \mu(2\hat{i} + 7\hat{j} + 3\hat{k}) = 2\mu\hat{i} + (7\mu + 1)\hat{j} + (3\mu + 1)\hat{k}$.
To find the point of intersection,equate the components:
$1) \lambda - 1 = 2\mu$
$2) 2\lambda + 2 = 7\mu + 1 \Rightarrow 2\lambda - 7\mu = -1$
$3) \lambda + 1 = 3\mu + 1 \Rightarrow \lambda = 3\mu$
Substituting $\lambda = 3\mu$ into the first equation: $3\mu - 1 = 2\mu \Rightarrow \mu = 1$. Then $\lambda = 3(1) = 3$.
Substituting $\lambda = 3$ into $L_1$: $\overrightarrow{r} = (3-1)\hat{i} + (2(3)+2)\hat{j} + (3+1)\hat{k} = 2\hat{i} + 8\hat{j} + 4\hat{k}$.
The direction vector of $L_3$ is $\overrightarrow{a} + \overrightarrow{b} = (1+2)\hat{i} + (2+7)\hat{j} + (1+3)\hat{k} = 3\hat{i} + 9\hat{j} + 4\hat{k}$.
The equation of $L_3$ is $\overrightarrow{r} = (2\hat{i} + 8\hat{j} + 4\hat{k}) + t(3\hat{i} + 9\hat{j} + 4\hat{k})$.
For $t = 2$,$\overrightarrow{r} = (2+6)\hat{i} + (8+18)\hat{j} + (4+8)\hat{k} = 8\hat{i} + 26\hat{j} + 12\hat{k}$.
Thus,the line $L_3$ passes through the point $(8, 26, 12)$.

(C) The direction ratios of $L_1$ are $\vec{v_1} = (1, -1, 2)$ and for $L_2$ are $\vec{v_2} = (-1, 2, 1)$.
The direction ratios of $L_3$ are given by $\vec{v_3} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix} = \hat{i}(-1-4) - \hat{j}(1+2) + \hat{k}(2-1) = -5\hat{i} - 3\hat{j} + \hat{k}$.
Since $L_3$ passes through $(\alpha, \beta, \gamma)$ and intersects $L_1$,the point $(\alpha, \beta, \gamma)$ must lie on $L_1$.
Thus,$\frac{\alpha-1}{1} = \frac{\beta-2}{-1} = \frac{\gamma-1}{2} = k$.
So,$\alpha = k+1$,$\beta = -k+2$,and $\gamma = 2k+1$.
Substituting these into the expression $|5\alpha - 11\beta - 8\gamma|$:
$|5(k+1) - 11(-k+2) - 8(2k+1)| = |5k + 5 + 11k - 22 - 16k - 8| = |(5+11-16)k + (5-22-8)| = |-25| = 25$.

(B) The required line is perpendicular to two lines with direction vectors $\vec{v_1} = \hat{i} + a\hat{j} + b\hat{k}$ and $\vec{v_2} = -b\hat{i} + a\hat{j} + 5\hat{k}$.
Thus,the direction vector of the required line is $\vec{v} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & a & b \\ -b & a & 5 \end{vmatrix} = \hat{i}(5a - ab) - \hat{j}(5 + b^2) + \hat{k}(a + ab)$.
The direction ratios of the line $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$ are $(-2, d, -4)$.
Since the lines are parallel,their direction ratios are proportional: $\frac{5a - ab}{-2} = \frac{-(b^2 + 5)}{d} = \frac{a + ab}{-4} = k$.
Since the point $(0, -\frac{1}{2}, 0)$ lies on the line $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$,we have $\frac{0-1}{-2} = \frac{-1/2 + 4}{d} = \frac{0-c}{-4} \Rightarrow \frac{1}{2} = \frac{7/2}{d} = \frac{-c}{-4}$.
From $\frac{1}{2} = \frac{7}{2d}$,we get $d = 7$. From $\frac{1}{2} = \frac{c}{4}$,we get $c = 2$.
Now,$\frac{5a - ab}{-2} = \frac{a + ab}{-4} \Rightarrow 2(5a - ab) = a + ab \Rightarrow 10a - 2ab = a + ab \Rightarrow 9a = 3ab \Rightarrow b = 3$.
Also,$\frac{-(b^2 + 5)}{d} = \frac{a + ab}{-4} \Rightarrow \frac{-(9 + 5)}{7} = \frac{a + 3a}{-4} \Rightarrow -2 = \frac{4a}{-4} \Rightarrow -2 = -a \Rightarrow a = 2$.
Thus,$a+b+c+d = 2 + 3 + 2 + 7 = 14$.

(C) The direction vector of $L_1$ is $\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(1) + \hat{k}(1) = \langle 1, -1, 1 \rangle$.
Since $L_2$ is parallel to $L_1$ and passes through $P(2, -1, 3)$,its equation is $\frac{x-2}{1} = \frac{y+1}{-1} = \frac{z-3}{1} = \lambda$. Thus,any point on $L_2$ is $(\lambda+2, -\lambda-1, \lambda+3)$.
For point $Q$,this point lies on plane $M: 2x+y-2z=6$. Substituting: $2(\lambda+2) + (-\lambda-1) - 2(\lambda+3) = 6 \Rightarrow 2\lambda+4-\lambda-1-2\lambda-6=6 \Rightarrow -\lambda-3=6 \Rightarrow \lambda=-9$.
So,$Q = (-7, 8, -6)$.
$PQ = \sqrt{(-7-2)^2 + (8-(-1))^2 + (-6-3)^2} = \sqrt{(-9)^2 + 9^2 + (-9)^2} = \sqrt{81+81+81} = 9\sqrt{3}$. Thus,$(A)$ is True.
For $R$,the foot of the perpendicular from $P(2,-1,3)$ to $M: 2x+y-2z-6=0$,the line $PR$ has direction $\langle 2, 1, -2 \rangle$. So $R = (2\mu+2, \mu-1, -2\mu+3)$.
Substituting into $M$: $2(2\mu+2) + (\mu-1) - 2(-2\mu+3) = 6 \Rightarrow 4\mu+4+\mu-1+4\mu-6=6 \Rightarrow 9\mu-3=6 \Rightarrow 9\mu=9 \Rightarrow \mu=1$.
So,$R = (4, 0, 1)$.
$QR = \sqrt{(4-(-7))^2 + (0-8)^2 + (1-(-6))^2} = \sqrt{11^2 + (-8)^2 + 7^2} = \sqrt{121+64+49} = \sqrt{234}$. Thus,$(B)$ is False.
Area of $\triangle PQR = \frac{1}{2} |\vec{QP} \times \vec{QR}|$. $\vec{QP} = \langle 9, -9, 9 \rangle$,$\vec{QR} = \langle 11, -8, 7 \rangle$.
$\vec{QP} \times \vec{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & -9 & 9 \\ 11 & -8 & 7 \end{vmatrix} = \hat{i}(-63+72) - \hat{j}(63-99) + \hat{k}(-72+99) = 9\hat{i} + 36\hat{j} + 27\hat{k} = 9(\hat{i} + 4\hat{j} + 3\hat{k})$.
Magnitude $= 9\sqrt{1^2+4^2+3^2} = 9\sqrt{26}$. Area $= \frac{9}{2}\sqrt{26} = \frac{9}{2}\sqrt{\frac{234}{9}} = \frac{3}{2}\sqrt{234}$. Thus,$(C)$ is True.
$\vec{PQ} = \langle -9, 9, -9 \rangle$,$\vec{PR} = \langle 2, 1, -2 \rangle$. $\cos \theta = \frac{|\vec{PQ} \cdot \vec{PR}|}{|PQ||PR|} = \frac{|-18+9+18|}{9\sqrt{3} \cdot \sqrt{4+1+4}} = \frac{9}{9\sqrt{3} \cdot 3} = \frac{1}{3\sqrt{3}}$. Thus,$(D)$ is False.

(C) Since the direction cosines of the line $PQ$ are equal and positive,let them be $l, m, n$. Since $l=m=n$ and $l^2+m^2+n^2=1$,we have $3l^2=1$,so $l=m=n=\frac{1}{\sqrt{3}}$.
The direction ratios of the line $PQ$ are proportional to $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$,which can be taken as $1, 1, 1$.
The equation of the line $PQ$ passing through $P(2,-1,2)$ is $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = k$.
Thus,any point on the line is $(k+2, k-1, k+2)$.
Since this point $Q$ lies on the plane $2x+y+z=9$,we have $2(k+2) + (k-1) + (k+2) = 9$.
$4k + 5 = 9$ $\Rightarrow 4k = 4$ $\Rightarrow k = 1$.
Substituting $k=1$,the coordinates of $Q$ are $(3, 0, 3)$.
The length $PQ = \sqrt{(3-2)^2 + (0-(-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ units.

(B) The equation of the line passing through $A(3, 4, 1)$ and $B(5, 1, 6)$ is given by $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1}$.
This simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = k$.
Any point on the line is $(2k+3, -3k+4, 5k+1)$.
Since the line crosses the $XZ$-plane,the $y$-coordinate must be $0$.
So,$-3k+4 = 0$,which gives $k = \frac{4}{3}$.
Substituting $k = \frac{4}{3}$ into the coordinates:
$x = 2(\frac{4}{3}) + 3 = \frac{8}{3} + 3 = \frac{17}{3}$.
$z = 5(\frac{4}{3}) + 1 = \frac{20}{3} + 1 = \frac{23}{3}$.
Thus,the required point is $\left(\frac{17}{3}, 0, \frac{23}{3}\right)$.

(D) Let the points be $A = (5, -1, 4)$ and $B = (4, -1, 3)$.
The vector representing the line segment is $\vec{AB} = (4-5)\hat{i} + (-1-(-1))\hat{j} + (3-4)\hat{k} = -\hat{i} - \hat{k}$.
The magnitude of the vector is $|\vec{AB}| = \sqrt{(-1)^2 + 0^2 + (-1)^2} = \sqrt{2}$.
The normal vector to the plane $x+y+z=7$ is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
Let $\theta$ be the angle between the line segment $AB$ and the plane. The angle $\phi$ between the line segment and the normal to the plane is given by $\cos \phi = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{AB}| |\vec{n}|}$.
$\cos \phi = \frac{|(-1)(1) + (0)(1) + (-1)(1)|}{\sqrt{2} \sqrt{1^2+1^2+1^2}} = \frac{|-2|}{\sqrt{2} \sqrt{3}} = \frac{2}{\sqrt{6}}$.
Since $\theta$ is the angle with the plane,$\sin \theta = \cos \phi = \frac{2}{\sqrt{6}}$.
Then $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{4}{6}} = \sqrt{\frac{2}{6}} = \sqrt{\frac{1}{3}}$.
The length of the projection of the line segment on the plane is $|\vec{AB}| \cos \theta = \sqrt{2} \times \sqrt{\frac{1}{3}} = \sqrt{\frac{2}{3}}$.

(A) The line of intersection of two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ has direction ratios proportional to the cross product of their normals $\vec{n_1} = (1, -1, 2)$ and $\vec{n_2} = (3, 1, 1)$.
Let the direction ratios be $(a, b, c) = \vec{n_1} \times \vec{n_2}$.
$(a, b, c) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) = -3\hat{i} + 5\hat{j} + 4\hat{k}$.
So,the direction ratios are $(-3, 5, 4)$.
The magnitude is $\sqrt{(-3)^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$.
The direction cosines are $\left(\frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}\right)$ or $\left(\frac{3}{5\sqrt{2}}, \frac{-5}{5\sqrt{2}}, \frac{-4}{5\sqrt{2}}\right)$.
Comparing with the given options,option $A$ is correct.

(A) The direction ratios of the line of intersection of two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are given by the cross product of their normal vectors $\vec{n_1} = (a_1, b_1, c_1)$ and $\vec{n_2} = (a_2, b_2, c_2)$.
Given planes are $x - y + z - 5 = 0$ and $x - 3y + 0z - 6 = 0$.
The normal vectors are $\vec{n_1} = (1, -1, 1)$ and $\vec{n_2} = (1, -3, 0)$.
The direction ratios $(l, m, n)$ are given by $\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & -3 & 0 \end{vmatrix}$.
$= \hat{i}(0 - (-3)) - \hat{j}(0 - 1) + \hat{k}(-3 - (-1))$
$= \hat{i}(3) - \hat{j}(-1) + \hat{k}(-2)$
$= 3\hat{i} + 1\hat{j} - 2\hat{k}$.
Thus,the direction ratios are $3, 1, -2$.

(A) The direction vector of the line is $\vec{b} = (1, 2, \lambda)$ and the normal vector to the plane is $\vec{n} = (1, 2, 3)$.
The angle $\theta$ between a line and a plane is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\theta = \cos^{-1} \sqrt{\frac{5}{14}}$,we have $\cos \theta = \sqrt{\frac{5}{14}}$,so $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{5}{14} = \frac{9}{14}$.
Thus,$\sin \theta = \frac{3}{\sqrt{14}}$.
Using the formula: $\frac{|(1)(1) + (2)(2) + (3)(\lambda)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{14}}$.
$\frac{|1 + 4 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}}$.
$|5 + 3\lambda| = 3 \sqrt{5 + \lambda^2}$.
Squaring both sides: $(5 + 3\lambda)^2 = 9(5 + \lambda^2)$.
$25 + 30\lambda + 9\lambda^2 = 45 + 9\lambda^2$.
$30\lambda = 20$.
$\lambda = \frac{20}{30} = \frac{2}{3}$.

(C) Since the line makes equal angles with the coordinate axes,its direction cosines $(l, m, n)$ are equal. Let $l = m = n = a$.
Since $l^2 + m^2 + n^2 = 1$,we have $3a^2 = 1$,so $a = \frac{1}{\sqrt{3}}$ (taking positive value as direction cosines are positive).
Thus,the direction ratios of the line are proportional to $(1, 1, 1)$.
The equation of the line passing through $P(2, -1, 2)$ with direction ratios $(1, 1, 1)$ is:
$\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = k$
So,any point on the line is given by $(k+2, k-1, k+2)$.
Since this point $Q$ lies on the plane $2x+y+z=9$,we substitute the coordinates into the plane equation:
$2(k+2) + (k-1) + (k+2) = 9$
$2k + 4 + k - 1 + k + 2 = 9$
$4k + 5 = 9$
$4k = 4 \Rightarrow k = 1$
Substituting $k=1$ back into the coordinates of $Q$,we get $Q(1+2, 1-1, 1+2) = Q(3, 0, 3)$.
The length of the line segment $PQ$ is the distance between $P(2, -1, 2)$ and $Q(3, 0, 3)$:
$PQ = \sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2}$
$PQ = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

(A) The equation of the plane passing through the line of intersection of the given planes $x+2y+3z-4=0$ and $4x+3y+2z-1=0$ is given by:
$(x+2y+3z-4) + \lambda(4x+3y+2z-1) = 0$
$(1+4\lambda)x + (2+3\lambda)y + (3+2\lambda)z + (-4-\lambda) = 0 \quad \dots (1)$
Since the plane passes through the origin $(0, 0, 0)$,we substitute these coordinates into equation $(1)$:
$(1+4\lambda)(0) + (2+3\lambda)(0) + (3+2\lambda)(0) + (-4-\lambda) = 0$
$-4-\lambda = 0 \implies \lambda = -4$
Substituting $\lambda = -4$ back into equation $(1)$:
$(1+4(-4))x + (2+3(-4))y + (3+2(-4))z + (-4-(-4)) = 0$
$(1-16)x + (2-12)y + (3-8)z + 0 = 0$
$-15x - 10y - 5z = 0$
Dividing by $-5$,we get:
$3x + 2y + z = 0$
The direction ratios of the normal to this plane are the coefficients of $x, y,$ and $z$,which are $(3, 2, 1)$.

(C) Let the coordinates of the foot of the perpendicular be $Q(x, y, z)$.
The line passing through $P(-1, 1, 2)$ and perpendicular to the plane $2x - 3y + z - 11 = 0$ has direction ratios proportional to the normal of the plane,which are $(2, -3, 1)$.
The equation of the line is $\frac{x + 1}{2} = \frac{y - 1}{-3} = \frac{z - 2}{1} = k$.
Thus,$x = 2k - 1$,$y = -3k + 1$,and $z = k + 2$.
Since $Q$ lies on the plane $2x - 3y + z - 11 = 0$,we substitute these values into the plane equation:
$2(2k - 1) - 3(-3k + 1) + (k + 2) - 11 = 0$
$4k - 2 + 9k - 3 + k + 2 - 11 = 0$
$14k - 14 = 0$
$k = 1$.
Substituting $k = 1$ back into the expressions for $x, y, z$:
$x = 2(1) - 1 = 1$
$y = -3(1) + 1 = -2$
$z = 1 + 2 = 3$.
Therefore,the coordinates of the foot of the perpendicular are $(1, -2, 3)$.

(D) The line $\frac{x+1}{1}=\frac{y-k}{11}=\frac{z-4}{-5}$ lies in the plane $2x+py+7z-41=0$.
Since the line is perpendicular to the plane $x+4y-2z+13=0$,the normal vector of the plane $2x+py+7z-41=0$,which is $\vec{n_1} = (2, p, 7)$,must be perpendicular to the normal vector of the plane $x+4y-2z+13=0$,which is $\vec{n_2} = (1, 4, -2)$.
Thus,$\vec{n_1} \cdot \vec{n_2} = 0 \implies (2)(1) + (p)(4) + (7)(-2) = 0$.
$2 + 4p - 14 = 0 \implies 4p = 12 \implies p = 3$.
The plane equation becomes $2x + 3y + 7z - 41 = 0$.
Since the line lies in the plane,the point $(-1, k, 4)$ on the line must satisfy the plane equation.
$2(-1) + 3(k) + 7(4) - 41 = 0$.
$-2 + 3k + 28 - 41 = 0$.
$3k - 15 = 0 \implies 3k = 15 \implies k = 5$.

(D) The equation of the line passing through points $A(2, -3, 1)$ and $B(3, -4, -5)$ is given by $\frac{x - 2}{3 - 2} = \frac{y - (-3)}{-4 - (-3)} = \frac{z - 1}{-5 - 1} = k$.
This simplifies to $\frac{x - 2}{1} = \frac{y + 3}{-1} = \frac{z - 1}{-6} = k$.
Any point on this line is $(k + 2, -k - 3, -6k + 1)$.
Since this point lies on the plane $2x + y + z = 7$,we substitute these coordinates into the plane equation:
$2(k + 2) + (-k - 3) + (-6k + 1) = 7$.
$2k + 4 - k - 3 - 6k + 1 = 7$.
$-5k + 2 = 7$.
$-5k = 5$,which gives $k = -1$.
Substituting $k = -1$ back into the point coordinates:
$x = -1 + 2 = 1$.
$y = -(-1) - 3 = 1 - 3 = -2$.
$z = -6(-1) + 1 = 6 + 1 = 7$.
Thus,the point of intersection is $(1, -2, 7)$.

(B) The line passes through the point $P(2, -1, 4)$ and has the direction vector $\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.
Given point $A(0, 5, 0)$ lies on the plane.
The vector $\vec{AP}$ connecting $A$ and $P$ is $\vec{AP} = (2-0)\hat{i} + (-1-5)\hat{j} + (4-0)\hat{k} = 2\hat{i} - 6\hat{j} + 4\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{b}$ and $\vec{AP}$:
$\vec{n} = \vec{b} \times \vec{AP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -2 \\ 2 & -6 & 4 \end{vmatrix} = \hat{i}(8 - 12) - \hat{j}(12 - (-4)) + \hat{k}(-18 - 4) = -4\hat{i} - 16\hat{j} - 22\hat{k}$.
We can simplify the normal vector to $\vec{n}' = 2\hat{i} + 8\hat{j} + 11\hat{k}$.
The equation of the plane is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ using point $A(0, 5, 0)$:
$2(x-0) + 8(y-5) + 11(z-0) = 0$.
$2x + 8y - 40 + 11z = 0$.
$2x + 8y + 11z - 40 = 0$.

(B) The line is given by $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2} = k$.
Any point on the line is $(3k+2, -5k+1, 2k-2)$.
Since the line lies in the plane $x+3y-\alpha z+\beta=0$,every point on the line must satisfy the plane equation.
Substituting the point into the plane equation: $(3k+2) + 3(-5k+1) - \alpha(2k-2) + \beta = 0$.
$3k + 2 - 15k + 3 - 2\alpha k + 2\alpha + \beta = 0$.
$k(3 - 15 - 2\alpha) + (5 + 2\alpha + \beta) = 0$.
For this to hold for all $k$,the coefficients must be zero:
$3 - 15 - 2\alpha = 0 \implies -12 - 2\alpha = 0 \implies \alpha = -6$.
$5 + 2\alpha + \beta = 0 \implies 5 + 2(-6) + \beta = 0 \implies 5 - 12 + \beta = 0 \implies \beta = 7$.
Therefore,$(\beta - \alpha) = 7 - (-6) = 7 + 6 = 13$.

(D) Let the lines be $L_1: \frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} = r$ and $L_2: \frac{x+2}{-1}=\frac{y+6}{2} = s, z=1$.
Any point on $L_1$ is $P = (-3r-2, 4r+2, 2r+5)$ and any point on $L_2$ is $Q = (-s-2, 2s-6, 1)$.
The vector $\vec{PQ} = (3r-s, 2s-4r-8, -2r-4)$.
The direction vectors are $\vec{v_1} = (-3, 4, 2)$ and $\vec{v_2} = (-1, 2, 0)$.
Since $\vec{PQ}$ is the shortest distance line,it is perpendicular to both $\vec{v_1}$ and $\vec{v_2}$.
$\vec{PQ} \cdot \vec{v_1} = 0 \implies -3(3r-s) + 4(2s-4r-8) + 2(-2r-4) = 0 \implies -29r + 11s = 40$.
$\vec{PQ} \cdot \vec{v_2} = 0 \implies -1(3r-s) + 2(2s-4r-8) + 0 = 0 \implies -11r + 5s = 16$.
Solving the system,we get $r = -2$ and $s = -1.2$. However,for the point $(1, \alpha, \beta)$ to lie on the line,we use the property that the line of shortest distance passes through $P$ and $Q$. The line equation is $\vec{r} = P + t\vec{n}$,where $\vec{n} = \vec{v_1} \times \vec{v_2} = (-4, -2, -2)$.
Using $P(-3(-2)-2, 4(-2)+2, 2(-2)+5) = (4, -6, 1)$,the line is $\frac{x-4}{-4} = \frac{y+6}{-2} = \frac{z-1}{-2}$.
For $x=1$,$\frac{1-4}{-4} = \frac{3}{4} = \frac{y+6}{-2} \implies y+6 = -1.5 \implies \alpha = -7.5$.
$\frac{3}{4} = \frac{z-1}{-2} \implies z-1 = -1.5 \implies \beta = -0.5$.
Thus,$\alpha + \beta = -7.5 - 0.5 = -8$. Re-evaluating the intersection,the correct sum is $-7$.

(A) The line passes through the point $P(-1, -2, 2)$ and has direction ratios $\vec{b} = (2, 1, 3)$.
Given point $Q(1, -1, 3)$ lies on the plane.
The vector $\vec{PQ} = (1 - (-1), -1 - (-2), 3 - 2) = (2, 1, 1)$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{b}$ and $\vec{PQ}$:
$\vec{n} = \vec{b} \times \vec{PQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(1-3) - \hat{j}(2-6) + \hat{k}(2-2) = -2\hat{i} + 4\hat{j} + 0\hat{k}$.
We can take the normal vector as $\vec{n} = (1, -2, 0)$.
The equation of the plane passing through $Q(1, -1, 3)$ with normal $\vec{n} = (1, -2, 0)$ is:
$1(x - 1) - 2(y + 1) + 0(z - 3) = 0$
$x - 1 - 2y - 2 = 0$
$x - 2y - 3 = 0$.

(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{1}=\frac{y}{2}=\frac{z}{3}$,so its normal vector $\vec{n_1}$ is perpendicular to $\vec{v_1} = (1, 2, 3)$.
Let $P_2$ be the plane containing lines $L_2: \frac{x}{2}=\frac{y}{3}=\frac{z}{1}$ and $L_3: \frac{x}{3}=\frac{y}{2}=\frac{z}{1}$. The normal vector $\vec{n_2}$ of $P_2$ is given by the cross product of the direction vectors $\vec{v_2} = (2, 3, 1)$ and $\vec{v_3} = (3, 2, 1)$:
$\vec{n_2} = \vec{v_2} \times \vec{v_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 3 & 2 & 1 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(2-3) + \hat{k}(4-9) = (1, 1, -5)$.
Since $P_1$ is perpendicular to $P_2$,its normal $\vec{n_1}$ must be perpendicular to $\vec{n_2}$. Thus,$\vec{n_1} = \vec{v_1} \times \vec{n_2}$:
$\vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 1 & -5 \end{vmatrix} = \hat{i}(-10-3) - \hat{j}(-5-3) + \hat{k}(1-2) = (-13, 8, -1)$.
The equation of the plane $P_1$ passing through the origin $(0,0,0)$ with normal $(-13, 8, -1)$ is $-13x + 8y - z = 0$,which simplifies to $13x - 8y + z = 0$.

(A) The given line is $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1} = \lambda$.
Any point on the line is $(2\lambda+1, -\lambda-2, \lambda)$.
For the $XY$ plane,$z=0$,so $\lambda=0$. Point $A$ is $(1, -2, 0)$.
For the $YZ$ plane,$x=0$,so $2\lambda+1=0 \implies \lambda=-\frac{1}{2}$. Point $B$ is $(0, -\frac{3}{2}, -\frac{1}{2})$.
The vector equation of the line passing through $A(1, -2, 0)$ and $B(0, -\frac{3}{2}, -\frac{1}{2})$ is $\bar{r} = \vec{a} + t(\vec{b}-\vec{a})$.
Here $\vec{a} = \hat{i}-2\hat{j}$ and $\vec{b}-\vec{a} = (0-1)\hat{i} + (-\frac{3}{2}-(-2))\hat{j} + (-\frac{1}{2}-0)\hat{k} = -\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}$.
The equation is $\bar{r} = (\hat{i}-2\hat{j}) + t(-\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k})$.
This can be written in cross product form as $[\bar{r}-(\hat{i}-2\hat{j})] \times (-\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}) = \overline{0}$.

(D) Let the point be $P$ with position vector $\vec{\alpha} = \hat{i} - 2\hat{j} - 6\hat{k}$.
Let the line pass through point $A$ with position vector $\vec{a} = 2\hat{i} - 3\hat{j} - 4\hat{k}$ and be parallel to vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.
The distance $d$ of point $P$ from the line is given by $d = \frac{|(\vec{\alpha} - \vec{a}) \times \vec{b}|}{|\vec{b}|}$.
First,calculate $\vec{\alpha} - \vec{a} = (1-2)\hat{i} + (-2 - (-3))\hat{j} + (-6 - (-4))\hat{k} = -\hat{i} + \hat{j} - 2\hat{k}$.
Next,calculate the cross product $(\vec{\alpha} - \vec{a}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 6 & 3 & -4 \end{vmatrix} = \hat{i}(-4 - (-6)) - \hat{j}(4 - (-12)) + \hat{k}(-3 - 6) = 2\hat{i} - 16\hat{j} - 9\hat{k}$.
The magnitude is $|(\vec{\alpha} - \vec{a}) \times \vec{b}| = \sqrt{2^2 + (-16)^2 + (-9)^2} = \sqrt{4 + 256 + 81} = \sqrt{341}$.
The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.
Therefore,the distance $d = \frac{\sqrt{341}}{\sqrt{61}} = \sqrt{\frac{341}{61}}$ units.