Line and Plane Questions in English

(C) The line makes equal angles with the coordinate axes,so its direction cosines are equal. Let the direction cosines be $(l, l, l)$. Since $l^2 + l^2 + l^2 = 1,$ we have $3l^2 = 1,$ so $l = \frac{1}{\sqrt{3}}$ (as direction cosines are positive).
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 $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = r.$
Any point $Q$ on this line is given by $(r+2, r-1, r+2).$
Since $Q$ lies on the plane $2x + y + z = 9,$ we substitute the coordinates of $Q$ into the plane equation:
$2(r+2) + (r-1) + (r+2) = 9$
$2r + 4 + r - 1 + r + 2 = 9$
$4r + 5 = 9$
$4r = 4 \Rightarrow r = 1.$
Thus,the point $Q$ is $(1+2, 1-1, 1+2) = (3, 0, 3).$
The distance $PQ$ is $\sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.$

(B) Let the direction ratios of the common line be $\ell, m, n$.
Since the line lies on the plane $x = 5$,its direction vector must be perpendicular to the normal $(1, 0, 0)$. Thus,$\ell = 0$.
Since the line lies on the other two planes,its direction vector must be perpendicular to their normals $(2, -5a, 3)$ and $(3b, 1, -3)$.
For the plane $2x - 5ay + 3z - 2 = 0$,we have $2\ell - 5am + 3n = 0$. Since $\ell = 0$,we get $-5am + 3n = 0$,or $3n = 5am$.
For the plane $3bx + y - 3z = 0$,we have $3b\ell + m - 3n = 0$. Since $\ell = 0$,we get $m - 3n = 0$,or $m = 3n$.
Substituting $m = 3n$ into $3n = 5am$,we get $3n = 5a(3n)$.
Assuming $n \neq 0$,we have $3 = 15a$,which gives $a = \frac{1}{5}$.
Now,the common line must also satisfy the equations of the planes. Since $x = 5$,the other two equations become $-5ay + 3z = 2 - 2(5) = -8$ and $y - 3z = -3b(5) = -15b$.
Substituting $a = \frac{1}{5}$,the first equation becomes $-y + 3z = -8$,or $y - 3z = 8$.
Comparing $y - 3z = 8$ and $y - 3z = -15b$,we get $-15b = 8$,so $b = -\frac{8}{15}$.
Thus,$(a, b) = \left( \frac{1}{5}, -\frac{8}{15} \right)$.

(A) The equation of a plane passing through the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is given by $a(x + 1) + b(y - 3) + c(z + 2) = 0$,where the normal vector $(a, b, c)$ is perpendicular to the line's direction vector $(-3, 2, 1)$.
Thus,$-3a + 2b + c = 0$ --- $(1)$
Since the plane passes through the point $(0, 7, -7)$,we substitute these coordinates into the plane equation:
$a(0 + 1) + b(7 - 3) + c(-7 + 2) = 0$
$a + 4b - 5c = 0$ --- $(2)$
Solving equations $(1)$ and $(2)$ using cross-multiplication:
$\frac{a}{2(-5) - 1(4)} = \frac{-b}{-3(-5) - 1(1)} = \frac{c}{-3(4) - 2(1)}$
$\frac{a}{-10 - 4} = \frac{-b}{15 - 1} = \frac{c}{-12 - 2}$
$\frac{a}{-14} = \frac{-b}{14} = \frac{c}{-14}$
Dividing by $-14$,we get $a = 1, b = 1, c = 1$.
Substituting these values into the plane equation:
$1(x + 1) + 1(y - 3) + 1(z + 2) = 0$
$x + y + z = 0$.

(C) Let the required line $L$ pass through $P(-4, 1, 3)$ with direction ratios $(a, b, c)$.
The equation of the line is $\frac{x + 4}{a} = \frac{y - 1}{b} = \frac{z - 3}{c}$.
Since $L$ is parallel to the plane $x + 2y - z - 5 = 0$,the normal to the plane $(1, 2, -1)$ is perpendicular to the line $L$. Thus,$a + 2b - c = 0$.
Let the line $L$ intersect the given line $L_1: \frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z - 2}{-1}$ at point $Q$. The vector $\vec{PQ}$ must be perpendicular to the normal of the plane and the direction of $L_1$. The vector $\vec{PQ}$ is $(x_Q - (-4), y_Q - 1, z_Q - 3)$.
Alternatively,for the line to intersect $L_1$ and be parallel to the plane,the direction vector $(a, b, c)$ must satisfy the condition that the scalar triple product of the direction of $L_1$ $(-3, 2, -1)$,the normal to the plane $(1, 2, -1)$,and the vector connecting the points $(-4, 1, 3)$ and $(-1, 3, 2)$ is zero.
The vector connecting the points is $(-1 - (-4), 3 - 1, 2 - 3) = (3, 2, -1)$.
The condition for intersection is $\begin{vmatrix} 3 & 2 & -1 \\ -3 & 2 & -1 \\ 1 & 2 & -1 \end{vmatrix} = 0$. Since this is not zero,we solve for $(a, b, c)$ using $a + 2b - c = 0$ and the condition that the line lies in the plane containing $L_1$ and parallel to the given plane.
Solving the system,we get the direction ratios $(3, -1, 1)$.
Thus,the equation is $\frac{x + 4}{3} = \frac{y - 1}{-1} = \frac{z - 3}{1}$.

(B) The equation of any plane passing through the intersection of the planes $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y + z - 4 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(x + y + z - 1) + \lambda(2x + 3y + z - 4) = 0$
$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 + \lambda)z - (1 + 4\lambda) = 0$.
Since the plane is parallel to the $y$-axis,the coefficient of $y$ must be zero.
$1 + 3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into the equation:
$(1 + 2(-\frac{1}{3}))x + (1 + 3(-\frac{1}{3}))y + (1 - \frac{1}{3})z - (1 + 4(-\frac{1}{3})) = 0$
$(1 - \frac{2}{3})x + 0y + \frac{2}{3}z - (1 - \frac{4}{3}) = 0$
$\frac{1}{3}x + \frac{2}{3}z + \frac{1}{3} = 0$
$x + 2z + 1 = 0$.
Checking the options for the point $(x, z)$ satisfying $x + 2z + 1 = 0$:
For option $A$: $(-3) + 2(-1) + 1 = -3 - 2 + 1 = -4 \neq 0$.
For option $B$: $(-3) + 2(1) + 1 = -3 + 2 + 1 = 0$.
Thus,the plane passes through $(-3, 1, 1)$.

(B) Let the first line be $L_1$ with direction vector $\vec{v_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Let the two lines in the second plane be $L_2$ and $L_3$ with direction vectors $\vec{v_2} = 3\hat{i} + 4\hat{j} + 2\hat{k}$ and $\vec{v_3} = 4\hat{i} + 2\hat{j} + 3\hat{k}$.
The normal vector $\vec{n_2}$ to the second plane is $\vec{v_2} \times \vec{v_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{vmatrix} = \hat{i}(12-4) - \hat{j}(9-8) + \hat{k}(6-16) = 8\hat{i} - \hat{j} - 10\hat{k}$.
The required plane contains $L_1$ and is perpendicular to the second plane,so its normal vector $\vec{n}$ must be perpendicular to both $\vec{v_1}$ and $\vec{n_2}$.
Thus,$\vec{n} = \vec{v_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 8 & -1 & -10 \end{vmatrix} = \hat{i}(-30+4) - \hat{j}(-20-32) + \hat{k}(-2-24) = -26\hat{i} + 52\hat{j} - 26\hat{k}$.
Taking the normal vector as $\hat{i} - 2\hat{j} + \hat{k}$,the equation of the plane passing through the origin $(0,0,0)$ is $1(x-0) - 2(y-0) + 1(z-0) = 0$,which simplifies to $x - 2y + z = 0$.

(A) The coordinates of point $A$ on the line are $(1 - 3\mu, \mu - 1, 2 + 5\mu)$.
The vector $\overrightarrow{AB}$ is given by $B - A = (3 - (1 - 3\mu))\hat{i} + (2 - (\mu - 1))\hat{j} + (6 - (2 + 5\mu))\hat{k}$.
$\overrightarrow{AB} = (2 + 3\mu)\hat{i} + (3 - \mu)\hat{j} + (4 - 5\mu)\hat{k}$.
The vector $\overrightarrow{AB}$ is parallel to the plane $x - 4y + 3z = 1$ if and only if the dot product of $\overrightarrow{AB}$ and the normal vector of the plane $\vec{n} = (1, -4, 3)$ is zero.
$(2 + 3\mu)(1) + (3 - \mu)(-4) + (4 - 5\mu)(3) = 0$.
$2 + 3\mu - 12 + 4\mu + 12 - 15\mu = 0$.
$(3 + 4 - 15)\mu + (2 - 12 + 12) = 0$.
$-8\mu + 2 = 0$.
$8\mu = 2 \Rightarrow \mu = \frac{2}{8} = \frac{1}{4}$.

(D) Let the given line be $L: \frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3} = k$. Any point on the line is $P(2k + 3, -k - 2, 3k + 1)$.
The plane $P_1$ is $2x + 3y - z = 5$. The normal vector to $P_1$ is $\vec{n_1} = 2\hat{i} + 3\hat{j} - \hat{k}$.
The direction of the line $L$ is $\vec{v} = 2\hat{i} - \hat{j} + 3\hat{k}$.
The plane $P_2$ contains the line $L$ and its projection on $P_1$. This means $P_2$ contains the line $L$ and is perpendicular to $P_1$.
The normal vector $\vec{n_2}$ of the required plane $P_2$ must be perpendicular to the direction of the line $\vec{v}$ and the normal of the plane $P_1$ $(\vec{n_1})$.
$\vec{n_2} = \vec{v} \times \vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 2 & 3 & -1 \end{vmatrix} = \hat{i}(1 - 9) - \hat{j}(-2 - 6) + \hat{k}(6 + 2) = -8\hat{i} + 8\hat{j} + 8\hat{k}$.
We can take the normal vector as $\vec{n} = \hat{i} - \hat{j} - \hat{k}$.
The plane passes through the point $(3, -2, 1)$ on the line. The equation of the plane is $1(x - 3) - 1(y + 2) - 1(z - 1) = 0$.
$x - 3 - y - 2 - z + 1 = 0 \Rightarrow x - y - z = 4$.
Checking the options:
$A: 2 - 2 - 0 = 0 \neq 4$
$B: -2 - 2 - 2 = -6 \neq 4$
$C: 0 - (-2) - 2 = 0 \neq 4$
$D: 2 - 0 - (-2) = 4$. Thus,point $(2, 0, -2)$ lies on the plane.

(B) The equation of the plane containing the two lines is given by the determinant form:
$\left| \begin{array}{ccc} x+2 & y-2 & z+5 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{array} \right| = 0$
Expanding the determinant:
$(x+2)(35-28) - (y-2)(21-7) + (z+5)(12-5) = 0$
$7(x+2) - 14(y-2) + 7(z+5) = 0$
Dividing by $7$:
$(x+2) - 2(y-2) + (z+5) = 0$
$x - 2y + z + 2 + 4 + 5 = 0$
$x - 2y + z + 11 = 0$
The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A=1, B=-2, C=1, D=11$.
$d = \frac{|11|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{11}{\sqrt{1+4+1}} = \frac{11}{\sqrt{6}}$.

(A) The direction ratios of the line are $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.
The normal vector to the plane is $\vec{n} = \hat{i} - 2\hat{j} - k\hat{k}$.
The angle $\alpha$ between a line and a plane is given by $\sin \alpha = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
$|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = 3$.
$|\vec{n}| = \sqrt{1^2 + (-2)^2 + (-k)^2} = \sqrt{5 + k^2}$.
$\vec{b} \cdot \vec{n} = (2)(1) + (1)(-2) + (-2)(-k) = 2 - 2 + 2k = 2k$.
So,$\sin \alpha = \frac{|2k|}{3\sqrt{5 + k^2}}$.
Given $\cos \alpha = \frac{2\sqrt{2}}{3}$,we find $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \frac{8}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Equating the two expressions for $\sin \alpha$:
$\frac{|2k|}{3\sqrt{5 + k^2}} = \frac{1}{3} \Rightarrow \frac{|2k|}{\sqrt{5 + k^2}} = 1$.
Squaring both sides: $\frac{4k^2}{5 + k^2} = 1$.
$4k^2 = 5 + k^2 \Rightarrow 3k^2 = 5 \Rightarrow k^2 = \frac{5}{3}$.
Thus,$k = \pm \sqrt{\frac{5}{3}}$. Given the options,the correct value is $\sqrt{\frac{5}{3}}$.

(D) The equation of a family of planes passing through the line of intersection of the planes $P_1: 2x - y - 4 = 0$ and $P_2: y + 2z - 4 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(2x - y - 4) + \lambda(y + 2z - 4) = 0$
Since the plane passes through the point $(1, 1, 0)$,we substitute $x = 1, y = 1, z = 0$ into the equation:
$(2(1) - 1 - 4) + \lambda(1 + 2(0) - 4) = 0$
$(2 - 1 - 4) + \lambda(1 - 4) = 0$
$-3 - 3\lambda = 0$
$-3\lambda = 3 \Rightarrow \lambda = -1$
Substituting $\lambda = -1$ back into the family equation:
$(2x - y - 4) - 1(y + 2z - 4) = 0$
$2x - y - 4 - y - 2z + 4 = 0$
$2x - 2y - 2z = 0$
Dividing by $2$,we get $x - y - z = 0$.

(C) The equation of the family of planes passing through the intersection of $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y + 4z - 5 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(x + y + z - 1) + \lambda(2x + 3y + 4z - 5) = 0$
$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z - (1 + 5\lambda) = 0$.
This plane is perpendicular to the plane $x - y + z = 0$. The condition for perpendicularity of two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Therefore,$(1 + 2\lambda)(1) + (1 + 3\lambda)(-1) + (1 + 4\lambda)(1) = 0$.
$1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$.
$1 + 3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into the equation of the plane:
$(1 + 2(-\frac{1}{3}))x + (1 + 3(-\frac{1}{3}))y + (1 + 4(-\frac{1}{3}))z - (1 + 5(-\frac{1}{3})) = 0$.
$(1 - \frac{2}{3})x + (1 - 1)y + (1 - \frac{4}{3})z - (1 - \frac{5}{3}) = 0$.
$\frac{1}{3}x + 0y - \frac{1}{3}z + \frac{2}{3} = 0$.
Multiplying by $3$,we get $x - z + 2 = 0$.
In vector form,this is $\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$.

(C) Any point on the given line can be represented as $(1 + 2\lambda, -1 + 3\lambda, 2 + 4\lambda)$ for some $\lambda \in \mathbb{R}$.
Substituting this point into the plane equation $x + 2y + 3z = 15$:
$(1 + 2\lambda) + 2(-1 + 3\lambda) + 3(2 + 4\lambda) = 15$
$1 + 2\lambda - 2 + 6\lambda + 6 + 12\lambda = 15$
$20\lambda + 5 = 15$
$20\lambda = 10$
$\lambda = \frac{1}{2}$
Substituting $\lambda = \frac{1}{2}$ back into the point coordinates:
$P = (1 + 2(\frac{1}{2}), -1 + 3(\frac{1}{2}), 2 + 4(\frac{1}{2})) = (2, \frac{1}{2}, 4)$.
The distance of $P(2, \frac{1}{2}, 4)$ from the origin $(0, 0, 0)$ is $\sqrt{2^2 + (\frac{1}{2})^2 + 4^2}$.
$= \sqrt{4 + \frac{1}{4} + 16} = \sqrt{20 + \frac{1}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$.

(D) The equation of any plane passing through the line of intersection of the planes $P_1: x + y + z - 6 = 0$ and $P_2: 2x + 3y + z + 5 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(x + y + z - 6) + \lambda(2x + 3y + z + 5) = 0$
$x(1 + 2\lambda) + y(1 + 3\lambda) + z(1 + \lambda) + (-6 + 5\lambda) = 0$.
Since the plane $P$ is perpendicular to the $xy$-plane (whose normal vector is $\vec{k} = (0, 0, 1)$),the normal vector of plane $P$,which is $\vec{n} = (1 + 2\lambda, 1 + 3\lambda, 1 + \lambda)$,must be perpendicular to $\vec{k}$.
Therefore,the dot product $\vec{n} \cdot \vec{k} = 0$,which implies $1 + \lambda = 0$,so $\lambda = -1$.
Substituting $\lambda = -1$ into the equation of the plane:
$x(1 - 2) + y(1 - 3) + z(1 - 1) + (-6 - 5) = 0$
$-x - 2y - 11 = 0$,or $x + 2y + 11 = 0$.
The distance of the point $(0, 0, 256)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
$d = \frac{|1(0) + 2(0) + 0(256) + 11|}{\sqrt{1^2 + 2^2 + 0^2}} = \frac{11}{\sqrt{1 + 4}} = \frac{11}{\sqrt{5}}$.

(A) Let $P$ be $(x_1, y_1, z_1)$. Since $Q(0, -1, -3)$ is the image of $P$ in the plane $3x - y + 4z - 2 = 0$,the line $PQ$ is perpendicular to the plane.
Direction ratios of the normal to the plane are $(3, -1, 4)$.
The line $PQ$ passes through $Q(0, -1, -3)$ and is parallel to the normal,so its equation is $\frac{x-0}{3} = \frac{y+1}{-1} = \frac{z+3}{4} = k$.
Any point on this line is $(3k, -k-1, 4k-3)$. The midpoint $M$ of $PQ$ lies on the plane.
$M = (\frac{3k}{2}, \frac{-k-2}{2}, \frac{4k-6}{2})$. Substituting into the plane equation: $3(\frac{3k}{2}) - (\frac{-k-2}{2}) + 4(\frac{4k-6}{2}) = 2$.
$9k + k + 2 + 16k - 24 = 4 \implies 26k = 26 \implies k = 1$.
Thus,$M = (1.5, -1, 1)$. The vector $\vec{QP} = 2\vec{QM} = 2(1.5, 0, 4) = (3, 0, 8)$.
$P = Q + (3, 0, 8) = (3, -1, 5)$.
Now,$\vec{QR} = (3-0, -1-(-1), -2-(-3)) = (3, 0, 1)$ and $\vec{QP} = (3, 0, 8)$.
Area of $\Delta PQR = \frac{1}{2} |\vec{QP} \times \vec{QR}|$.
$\vec{QP} \times \vec{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 8 \\ 3 & 0 & 1 \end{vmatrix} = \hat{i}(0) - \hat{j}(3-24) + \hat{k}(0) = 21\hat{j}$.
Area $= \frac{1}{2} |21\hat{j}| = \frac{21}{2} = 10.5$.
Wait,re-evaluating the provided solution logic: The distance from $R$ to line $PQ$ is needed. The area is $\frac{1}{2} \times \text{base} \times \text{height}$. Base $PQ = \sqrt{3^2 + 0^2 + 8^2} = \sqrt{73}$. Height is the perpendicular distance from $R(3, -1, -2)$ to line $PQ$.
Using the cross product method: Area $= \frac{1}{2} |\vec{QP} \times \vec{QR}| = \frac{21}{2} = 10.5$.

(A) Let a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1} = \lambda$ be $(2\lambda + 1, -\lambda - 1, \lambda)$.
Since the foot of the perpendicular $Q$ lies on both planes $x + y + z = 3$ and $x - y + z = 3$,we can find the intersection of these two planes.
Adding the two equations: $(x + y + z) + (x - y + z) = 3 + 3 \Rightarrow 2x + 2z = 6 \Rightarrow x + z = 3$.
Subtracting the two equations: $(x + y + z) - (x - y + z) = 3 - 3 \Rightarrow 2y = 0 \Rightarrow y = 0$.
Since $Q$ lies on the line of intersection of the two planes,its coordinates must satisfy $y = 0$ and $z = 3 - x$.
Thus,$Q$ is of the form $(x, 0, 3 - x)$.
Since $PQ$ is perpendicular to the plane $x + y + z = 3$,the vector $\vec{PQ}$ must be parallel to the normal vector of the plane,$\vec{n} = (1, 1, 1)$.
$vec{PQ} = (x - (2\lambda + 1), 0 - (-\lambda - 1), 3 - x - \lambda) = (x - 2\lambda - 1, \lambda + 1, 3 - x - \lambda)$.
Since $\vec{PQ} = k(1, 1, 1)$,we have $x - 2\lambda - 1 = \lambda + 1 = 3 - x - \lambda$.
From $\lambda + 1 = 3 - x - \lambda$,we get $2\lambda + x = 2$.
From $x - 2\lambda - 1 = \lambda + 1$,we get $x - 3\lambda = 2$.
Solving $2\lambda + x = 2$ and $x - 3\lambda = 2$,we subtract the equations: $5\lambda = 0 \Rightarrow \lambda = 0$.
Substituting $\lambda = 0$ into $x - 3\lambda = 2$,we get $x = 2$.
Thus,the coordinates of $Q$ are $(2, 0, 3 - 2) = (2, 0, 1)$.

(A) Let the line be $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=\lambda$.
Then,any point on the line is given by $x=3\lambda+2, y=2\lambda-1, z=-\lambda+1$.
For the intersection with the plane $2x+3y-z+13=0$:
$2(3\lambda+2)+3(2\lambda-1)-(-\lambda+1)+13=0$
$6\lambda+4+6\lambda-3+\lambda-1+13=0$
$13\lambda+13=0 \implies \lambda=-1$.
Substituting $\lambda=-1$,we get $P(-1, -3, 2)$.
For the intersection with the plane $3x+y+4z=16$:
$3(3\lambda+2)+(2\lambda-1)+4(-\lambda+1)=16$
$9\lambda+6+2\lambda-1-4\lambda+4=16$
$7\lambda+9=16 \implies 7\lambda=7 \implies \lambda=1$.
Substituting $\lambda=1$,we get $Q(5, 1, 0)$.
The distance $PQ$ is given by $\sqrt{(5 - (-1))^2 + (1 - (-3))^2 + (0 - 2)^2}$.
$PQ = \sqrt{6^2 + 4^2 + (-2)^2} = \sqrt{36 + 16 + 4} = \sqrt{56} = 2\sqrt{14}$.

(A) The equation of a plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by $\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2.$
Given planes are $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5.$
Thus,the equation of the required plane is $\vec{r} \cdot [(\hat{i}+\hat{j}+\hat{k}) + \lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})] = 6 - 5\lambda.$
This can be written as $\vec{r} \cdot [(1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}] = 6 - 5\lambda.$
Since the plane passes through the point $(1,1,1),$ we substitute $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ into the equation:
$(1+2\lambda)(1) + (1+3\lambda)(1) + (1+4\lambda)(1) = 6 - 5\lambda.$
$1 + 2\lambda + 1 + 3\lambda + 1 + 4\lambda = 6 - 5\lambda.$
$3 + 9\lambda = 6 - 5\lambda.$
$14\lambda = 3 \implies \lambda = \frac{3}{14}.$
Substituting $\lambda = \frac{3}{14}$ back into the equation:
$\vec{r} \cdot [(1 + 2(\frac{3}{14}))\hat{i} + (1 + 3(\frac{3}{14}))\hat{j} + (1 + 4(\frac{3}{14}))\hat{k}] = 6 - 5(\frac{3}{14}).$
$\vec{r} \cdot [(1 + \frac{3}{7})\hat{i} + (1 + \frac{9}{14})\hat{j} + (1 + \frac{6}{7})\hat{k}] = \frac{84 - 15}{14}.$
$\vec{r} \cdot [\frac{10}{7}\hat{i} + \frac{23}{14}\hat{j} + \frac{13}{7}\hat{k}] = \frac{69}{14}.$
Multiplying by $14,$ we get $\vec{r} \cdot (20\hat{i} + 23\hat{j} + 26\hat{k}) = 69.$

(N/A) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
From the given equations:
$(x_1, y_1, z_1) = (-3, 1, 5)$ and $(a_1, b_1, c_1) = (-3, 1, 5)$.
$(x_2, y_2, z_2) = (-1, 2, 5)$ and $(a_2, b_2, c_2) = (-1, 2, 5)$.
Now,calculate the determinant:
$\begin{vmatrix} -1-(-3) & 2-1 & 5-5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$.
Expanding along the first row:
$= 2(1 \times 5 - 5 \times 2) - 1(-3 \times 5 - 5 \times -1) + 0(-3 \times 2 - 1 \times -1)$
$= 2(5 - 10) - 1(-15 + 5) + 0$
$= 2(-5) - 1(-10) = -10 + 10 = 0$.
Since the determinant is $0$,the lines are coplanar.

(A) The direction vector of the line is $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
The normal vector to the plane is $\vec{n} = 10\hat{i} + 2\hat{j} - 11\hat{k}$.
Let $\phi$ be the angle between the line and the plane. The formula for the angle is given by $\sin \phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
First,calculate the dot product: $\vec{b} \cdot \vec{n} = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = -40$.
Next,calculate the magnitudes: $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$|\vec{n}| = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$.
Thus,$\sin \phi = \frac{|-40|}{7 \times 15} = \frac{40}{105} = \frac{8}{21}$.
Therefore,$\phi = \sin^{-1}\left(\frac{8}{21}\right)$.

(A) The equations of the planes are $\vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) = 7$ and $\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9$.
The equation of any plane passing through the intersection of these two planes is given by:
$[\vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) - 7] + \lambda [\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9] = 0$,where $\lambda \in \mathbb{R}$.
Rearranging the terms,we get:
$\vec{r} \cdot [(2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k}] = 7 + 9\lambda$ ... $(1)$
Since the plane passes through the point $(2,1,3)$,its position vector is $\vec{r} = 2\hat{i} + \hat{j} + 3\hat{k}$.
Substituting this into equation $(1)$:
$(2\hat{i} + \hat{j} + 3\hat{k}) \cdot [(2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (3\lambda - 3)\hat{k}] = 7 + 9\lambda$
Calculating the dot product:
$2(2 + 2\lambda) + 1(2 + 5\lambda) + 3(3\lambda - 3) = 7 + 9\lambda$
$4 + 4\lambda + 2 + 5\lambda + 9\lambda - 9 = 7 + 9\lambda$
$18\lambda - 3 = 7 + 9\lambda$
$9\lambda = 10 \Rightarrow \lambda = \frac{10}{9}$.
Substituting $\lambda = \frac{10}{9}$ into equation $(1)$:
$\vec{r} \cdot [(2 + 2(\frac{10}{9}))\hat{i} + (2 + 5(\frac{10}{9}))\hat{j} + (3(\frac{10}{9}) - 3)\hat{k}] = 7 + 9(\frac{10}{9})$
$\vec{r} \cdot [(\frac{18+20}{9})\hat{i} + (\frac{18+50}{9})\hat{j} + (\frac{30-27}{9})\hat{k}] = 7 + 10$
$\vec{r} \cdot [\frac{38}{9}\hat{i} + \frac{68}{9}\hat{j} + \frac{3}{9}\hat{k}] = 17$
Multiplying by $9$,we get:
$\vec{r} \cdot (38\hat{i} + 68\hat{j} + 3\hat{k}) = 153$.

(A) The equation of the plane passing through the intersection of the planes $x+y+z-1=0$ and $2x+3y+4z-5=0$ is given by:
$(x+y+z-1) + \lambda(2x+3y+4z-5) = 0$
$(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z - (1+5\lambda) = 0$ ... $(1)$
The normal vector to this plane is $\vec{n_1} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}$.
The plane is perpendicular to the plane $x-y+z=0$,which has a normal vector $\vec{n_2} = 1\hat{i} - 1\hat{j} + 1\hat{k}$.
Since the planes are perpendicular,their normal vectors are perpendicular,so $\vec{n_1} \cdot \vec{n_2} = 0$:
$(1+2\lambda)(1) + (1+3\lambda)(-1) + (1+4\lambda)(1) = 0$
$1+2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$
$3\lambda + 1 = 0$
$\lambda = -\frac{1}{3}$
Substituting $\lambda = -\frac{1}{3}$ into equation $(1)$:
$(1+2(-\frac{1}{3}))x + (1+3(-\frac{1}{3}))y + (1+4(-\frac{1}{3}))z - (1+5(-\frac{1}{3})) = 0$
$(\frac{1}{3})x + (0)y + (-\frac{1}{3})z - (-\frac{2}{3}) = 0$
$\frac{1}{3}x - \frac{1}{3}z + \frac{2}{3} = 0$
Multiplying by $3$,we get $x-z+2=0$.

(C) The distance $d$ of a point $P(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula:
$d = \left| \frac{Ax_1 + By_1 + Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}} \right|$
Given the point $(x_1, y_1, z_1) = (2, 3, -5)$ and the plane equation $x + 2y - 2z - 9 = 0$,we have $A = 1, B = 2, C = -2, D = -9$.
Substituting these values into the formula:
$d = \left| \frac{1(2) + 2(3) - 2(-5) - 9}{\sqrt{1^2 + 2^2 + (-2)^2}} \right|$
$d = \left| \frac{2 + 6 + 10 - 9}{\sqrt{1 + 4 + 4}} \right|$
$d = \left| \frac{9}{\sqrt{9}} \right|$
$d = \frac{9}{3} = 3$
Thus,the distance is $3$ units.

(A) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
For the given lines,we have:
$(x_1, y_1, z_1) = (a-d, a, a+d)$ and $(a_1, b_1, c_1) = (\alpha-\delta, \alpha, \alpha+\delta)$.
$(x_2, y_2, z_2) = (b-c, b, b+c)$ and $(a_2, b_2, c_2) = (\beta-\gamma, \beta, \beta+\gamma)$.
Now,consider the determinant:
$\Delta = \begin{vmatrix} b-c-a+d & b-a & b+c-a-d \\ \alpha-\delta & \alpha & \alpha+\delta \\ \beta-\gamma & \beta & \beta+\gamma \end{vmatrix}$.
Applying the column operation $C_1 \to C_1 + C_3$:
$\Delta = \begin{vmatrix} (b-c-a+d) + (b+c-a-d) & b-a & b+c-a-d \\ (\alpha-\delta) + (\alpha+\delta) & \alpha & \alpha+\delta \\ (\beta-\gamma) + (\beta+\gamma) & \beta & \beta+\gamma \end{vmatrix} = \begin{vmatrix} 2(b-a) & b-a & b+c-a-d \\ 2\alpha & \alpha & \alpha+\delta \\ 2\beta & \beta & \beta+\gamma \end{vmatrix}$.
Taking $2$ as a common factor from the first column:
$\Delta = 2 \begin{vmatrix} b-a & b-a & b+c-a-d \\ \alpha & \alpha & \alpha+\delta \\ \beta & \beta & \beta+\gamma \end{vmatrix}$.
Since the first and second columns are identical,the value of the determinant is $0$.
Thus,the given lines are coplanar.

(A) The position vector of the point $(1, 2, 3)$ is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The plane is given by $\vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0$. The normal vector to the plane is $\vec{n} = \hat{i} + 2\hat{j} - 5\hat{k}$.
$A$ line perpendicular to the plane will have the same direction as the normal vector of the plane. Thus,the direction vector of the line is $\vec{v} = \hat{i} + 2\hat{j} - 5\hat{k}$.
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to vector $\vec{v}$ is $\vec{r} = \vec{a} + \lambda\vec{v}$.
Substituting the values,we get $\vec{l} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} - 5\hat{k})$.

(A) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
Substituting the given points $(3, -4, -5)$ and $(2, -3, 1)$,we get:
$\frac{x-3}{2-3} = \frac{y+4}{-3+4} = \frac{z+5}{1+5}$
$\Rightarrow \frac{x-3}{-1} = \frac{y+4}{1} = \frac{z+5}{6} = k$ (say).
Expressing $x, y, z$ in terms of $k$:
$x = 3 - k, y = k - 4, z = 6k - 5$.
Since this point lies on the plane $2x + y + z = 7$,we substitute these values into the plane equation:
$2(3 - k) + (k - 4) + (6k - 5) = 7$
$6 - 2k + k - 4 + 6k - 5 = 7$
$5k - 3 = 7$
$5k = 10 \Rightarrow k = 2$.
Substituting $k = 2$ back into the coordinates:
$x = 3 - 2 = 1$
$y = 2 - 4 = -2$
$z = 6(2) - 5 = 7$.
Thus,the required point is $(1, -2, 7)$.

(A) The perpendicular distance $D$ from a point with position vector $\vec{a}$ to the plane $\vec{r} \cdot \vec{n} + d = 0$ is given by $D = \frac{|\vec{a} \cdot \vec{n} + d|}{|\vec{n}|}$.
For the given plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0$,we have $\vec{n} = 3 \hat{i} + 4 \hat{j} - 12 \hat{k}$ and $d = 13$.
The magnitude $|\vec{n}| = \sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.
Let $D_1$ be the distance from $(1, 1, p)$ to the plane:
$D_1 = \frac{|(1)(3) + (1)(4) + (p)(-12) + 13|}{13} = \frac{|3 + 4 - 12p + 13|}{13} = \frac{|20 - 12p|}{13}$.
Let $D_2$ be the distance from $(-3, 0, 1)$ to the plane:
$D_2 = \frac{|(-3)(3) + (0)(4) + (1)(-12) + 13|}{13} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.
Since the points are equidistant,$D_1 = D_2$:
$\frac{|20 - 12p|}{13} = \frac{8}{13} \implies |20 - 12p| = 8$.
Case $1$: $20 - 12p = 8 \implies 12p = 12 \implies p = 1$.
Case $2$: $20 - 12p = -8 \implies 12p = 28 \implies p = \frac{28}{12} = \frac{7}{3}$.
Thus,the values of $p$ are $1$ and $\frac{7}{3}$.

(A) The given planes are $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1=0$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$.
The equation of any plane passing through the line of intersection of these planes is given by:
$[\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1] + \lambda[\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4] = 0$
Rearranging the terms,we get:
$\vec{r} \cdot[(1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1-\lambda)\hat{k}] + (4\lambda-1) = 0$ ..........$(1)$
The normal vector to this plane is $\vec{n} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1-\lambda)\hat{k}$.
Since the required plane is parallel to the $x$-axis,its normal vector must be perpendicular to the $x$-axis. The direction vector of the $x$-axis is $\hat{i} = (1, 0, 0)$.
Thus,$\vec{n} \cdot \hat{i} = 0$:
$(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$
$1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into equation $(1)$:
$\vec{r} \cdot[(1-1)\hat{i} + (1-\frac{3}{2})\hat{j} + (1+\frac{1}{2})\hat{k}] + (4(-\frac{1}{2})-1) = 0$
$\vec{r} \cdot[0\hat{i} - \frac{1}{2}\hat{j} + \frac{3}{2}\hat{k}] - 3 = 0$
Multiplying by $-2$:
$\vec{r} \cdot[\hat{j} - 3\hat{k}] + 6 = 0$.
In Cartesian form,where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,this becomes:
$y - 3z + 6 = 0$.

(A) The equations of the given planes are:
$\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$ $(1)$
$\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ $(2)$
The equation of the plane passing through the line of intersection of planes $(1)$ and $(2)$ is given by:
$[\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4] + \lambda[\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5] = 0$
$\vec{r} \cdot[(1+2\lambda) \hat{i} + (2+\lambda) \hat{j} + (3-\lambda) \hat{k}] + (5\lambda-4) = 0$ $(3)$
Since this plane is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$,the dot product of their normal vectors is zero:
$5(1+2\lambda) + 3(2+\lambda) - 6(3-\lambda) = 0$
$5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0$
$19\lambda - 7 = 0 \Rightarrow \lambda = \frac{7}{19}$
Substituting $\lambda = \frac{7}{19}$ into equation $(3)$:
$\vec{r} \cdot[(1 + \frac{14}{19}) \hat{i} + (2 + \frac{7}{19}) \hat{j} + (3 - \frac{7}{19}) \hat{k}] + (5(\frac{7}{19}) - 4) = 0$
$\vec{r} \cdot[\frac{33}{19} \hat{i} + \frac{45}{19} \hat{j} + \frac{50}{19} \hat{k}] + (\frac{35-76}{19}) = 0$
$\vec{r} \cdot(33 \hat{i} + 45 \hat{j} + 50 \hat{k}) - 41 = 0$

(A) The equation of the given line is $\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})$ .........$(1)$
The equation of the given plane is $\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$ .........$(2)$
Substituting the value of $\vec{r}$ from equation $(1)$ in equation $(2)$,we obtain:
$[(3 \lambda+2) \hat{i}+(4 \lambda-1) \hat{j}+(2 \lambda+2) \hat{k}] \cdot(\hat{i}-\hat{j}+\hat{k})=5$
$(3 \lambda+2)-(4 \lambda-1)+(2 \lambda+2)=5$
$3 \lambda+2-4 \lambda+1+2 \lambda+2=5$
$\lambda+5=5$
$\lambda=0$
Substituting $\lambda=0$ in equation $(1)$,the point of intersection is $(2, -1, 2)$.
The distance $d$ between the points $(2, -1, 2)$ and $(-1, -5, -10)$ is:
$d=\sqrt{(-1-2)^{2}+(-5-(-1))^{2}+(-10-2)^{2}}$
$d=\sqrt{(-3)^{2}+(-4)^{2}+(-12)^{2}}$
$d=\sqrt{9+16+144}$
$d=\sqrt{169}$
$d=13$

(A) Let the required line be parallel to vector $\vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$.
The position vector of the point $(1, 2, 3)$ is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The equation of the line passing through $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r} = \vec{a} + \lambda\vec{b}$.
Since the line is parallel to the given planes,the vector $\vec{b}$ must be perpendicular to the normals of both planes.
The normals are $\vec{n}_{1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n}_{2} = 3\hat{i} + \hat{j} + \hat{k}$.
Thus,$\vec{b} = \vec{n}_{1} \times \vec{n}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) = -3\hat{i} + 5\hat{j} + 4\hat{k}$.
Substituting $\vec{a}$ and $\vec{b}$ into the line equation,we get $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-3\hat{i} + 5\hat{j} + 4\hat{k})$.

(A) The distance of a point with position vector $\vec{a}$ from the plane $\vec{r} \cdot \vec{n} = d$ is given by the formula: $\text{Distance} = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$.
Given $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$,$\vec{n} = \hat{i} - 2 \hat{j} + 4 \hat{k}$,and $d = 9$.
First,calculate the dot product $\vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4$.
Next,calculate the magnitude of the normal vector $|\vec{n}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Substitute these values into the formula:
$\text{Distance} = \frac{|-4 - 9|}{\sqrt{21}} = \frac{|-13|}{\sqrt{21}} = \frac{13}{\sqrt{21}}$.

(A) The equation of the plane passing through the points $A(2, 2, 1)$,$B(3, 0, 1)$,and $C(4, -1, 0)$ is given by the determinant form:
$\begin{vmatrix} x-2 & y-2 & z-1 \\ 3-2 & 0-2 & 1-1 \\ 4-2 & -1-2 & 0-1 \end{vmatrix} = 0$
$\begin{vmatrix} x-2 & y-2 & z-1 \\ 1 & -2 & 0 \\ 2 & -3 & -1 \end{vmatrix} = 0$
$(x-2)(2-0) - (y-2)(-1-0) + (z-1)(-3+4) = 0$
$2(x-2) + 1(y-2) + 1(z-1) = 0$
$2x - 4 + y - 2 + z - 1 = 0$
$2x + y + z - 7 = 0 \ldots(1)$
The equation of the line passing through $(3, -4, -5)$ and $(2, -3, 1)$ is:
$\frac{x-3}{2-3} = \frac{y+4}{-3+4} = \frac{z+5}{1+5} = \lambda$
$\frac{x-3}{-1} = \frac{y+4}{1} = \frac{z+5}{6} = \lambda$
Any point on this line is given by $(x, y, z) = (-\lambda+3, \lambda-4, 6\lambda-5)$.
Since this point lies on the plane $(1)$,we substitute these coordinates into the plane equation:
$2(-\lambda+3) + (\lambda-4) + (6\lambda-5) - 7 = 0$
$-2\lambda + 6 + \lambda - 4 + 6\lambda - 5 - 7 = 0$
$5\lambda - 10 = 0 \implies 5\lambda = 10 \implies \lambda = 2$
Substituting $\lambda = 2$ back into the point coordinates:
$x = -2+3 = 1$
$y = 2-4 = -2$
$z = 6(2)-5 = 7$
Thus,the required point is $(1, -2, 7)$.

(A) The equation of the line is $\vec{r} = (2 + 3\lambda)\hat{i} + (-1 + 4\lambda)\hat{j} + (2 + 2\lambda)\hat{k}$.
Substituting this into the plane equation $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$:
$((2 + 3\lambda)\hat{i} + (-1 + 4\lambda)\hat{j} + (2 + 2\lambda)\hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$
$(2 + 3\lambda) - (-1 + 4\lambda) + (2 + 2\lambda) = 5$
$2 + 3\lambda + 1 - 4\lambda + 2 + 2\lambda = 5$
$5 + \lambda = 5 \implies \lambda = 0$.
Substituting $\lambda = 0$ into the line equation,the point of intersection is $(2, -1, 2)$.
The distance between $(2, -1, 2)$ and $(-1, -5, -10)$ is given by the distance formula:
$d = \sqrt{(-1 - 2)^2 + (-5 - (-1))^2 + (-10 - 2)^2}$
$d = \sqrt{(-3)^2 + (-4)^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

(N/A) Let the given point be $P(\hat{i}+3 \hat{j}+4 \hat{k})$ and $Q$ be the image of $P$ in the plane $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$.
Then $PQ$ is the normal to the plane. Since $PQ$ passes through $P$ and is normal to the given plane,the equation of line $PQ$ is given by $\vec{r}=(\hat{i}+3 \hat{j}+4 \hat{k})+\lambda(2 \hat{i}-\hat{j}+\hat{k})$.
Since $Q$ lies on the line $PQ$,the position vector of $Q$ can be expressed as $(1+2 \lambda) \hat{i}+(3-\lambda) \hat{j}+(4+\lambda) \hat{k}$.
Let $R$ be the point of intersection of the line $PQ$ and the plane. Since $R$ is the midpoint of $PQ$,the position vector of $R$ is $\frac{[(1+2 \lambda) \hat{i}+(3-\lambda) \hat{j}+(4+\lambda) \hat{k}]+[\hat{i}+3 \hat{j}+4 \hat{k}]}{2} = (1+\lambda) \hat{i} + (3-\frac{\lambda}{2}) \hat{j} + (4+\frac{\lambda}{2}) \hat{k}$.
Since $R$ lies on the plane $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$,we have:
$[(1+\lambda) \hat{i} + (3-\frac{\lambda}{2}) \hat{j} + (4+\frac{\lambda}{2}) \hat{k}] \cdot (2 \hat{i}-\hat{j}+\hat{k}) + 3 = 0$
$2(1+\lambda) - (3-\frac{\lambda}{2}) + (4+\frac{\lambda}{2}) + 3 = 0$
$2+2\lambda - 3 + \frac{\lambda}{2} + 4 + \frac{\lambda}{2} + 3 = 0$
$3\lambda + 6 = 0 \Rightarrow \lambda = -2$.
Substituting $\lambda = -2$ into the expression for $Q$:
$Q = (1+2(-2)) \hat{i} + (3-(-2)) \hat{j} + (4+(-2)) \hat{k}$
$Q = -3 \hat{i} + 5 \hat{j} + 2 \hat{k}$.

(A) The equation of the given plane is $2x - 2y + 4z + 5 = 0$ ... $(i)$
The normal vector to the plane is $\vec{n} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.
The equation of the line passing through the point $P\left(1, \frac{3}{2}, 2\right)$ and perpendicular to the plane is given by $\frac{x-1}{2} = \frac{y-3/2}{-2} = \frac{z-2}{4} = \lambda$.
Thus,any point on this line is given by $x = 2\lambda + 1$,$y = -2\lambda + \frac{3}{2}$,and $z = 4\lambda + 2$.
If this point lies on the plane,it must satisfy the plane equation:
$2(2\lambda + 1) - 2(-2\lambda + \frac{3}{2}) + 4(4\lambda + 2) + 5 = 0$
$4\lambda + 2 + 4\lambda - 3 + 16\lambda + 8 + 5 = 0$
$24\lambda + 12 = 0 \Rightarrow 24\lambda = -12 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the coordinates,the foot of the perpendicular is $\left(2(-\frac{1}{2}) + 1, -2(-\frac{1}{2}) + \frac{3}{2}, 4(-\frac{1}{2}) + 2\right) = \left(0, \frac{5}{2}, 0\right)$.
The length of the perpendicular is the distance between $\left(1, \frac{3}{2}, 2\right)$ and $\left(0, \frac{5}{2}, 0\right)$:
$d = \sqrt{(1-0)^2 + (\frac{3}{2} - \frac{5}{2})^2 + (2-0)^2} = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$ units.

(A) The equations of the two planes are $x+2y=0$ and $3y-z=0$.
Let $\vec{n}_{1}$ and $\vec{n}_{2}$ be the normals to the two planes,respectively.
$\vec{n}_{1} = \hat{i} + 2\hat{j} + 0\hat{k}$ and $\vec{n}_{2} = 0\hat{i} + 3\hat{j} - \hat{k}$.
Since the required line is parallel to both planes,it must be parallel to the cross product of their normals,$\vec{b} = \vec{n}_{1} \times \vec{n}_{2}$.
$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 0 & 3 & -1 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(-1 - 0) + \hat{k}(3 - 0) = -2\hat{i} + \hat{j} + 3\hat{k}$.
The line passes through the point $(3, 0, 1)$.
The equation of the line in symmetric form is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Substituting the point $(3, 0, 1)$ and the direction vector $(-2, 1, 3)$,we get $\frac{x-3}{-2} = \frac{y-0}{1} = \frac{z-1}{3}$.

(D) Let the equation of the plane be $a(x - 2) + b(y - 1) + c(z + 1) = 0$ ... $(i)$
Since it passes through $(-1, 3, 4)$,we have $a(-1 - 2) + b(3 - 1) + c(4 + 1) = 0$,which simplifies to $-3a + 2b + 5c = 0$ ... $(ii)$
The plane $(i)$ is perpendicular to $x - 2y + 4z = 10$,so the normal vectors are perpendicular: $1(a) - 2(b) + 4(c) = 0$,which gives $a - 2b + 4c = 0$ ... $(iii)$
Solving $(ii)$ and $(iii)$ using cross multiplication:
$\frac{a}{(2)(4) - (5)(-2)} = \frac{-b}{(-3)(4) - (5)(1)} = \frac{c}{(-3)(-2) - (2)(1)}$
$\frac{a}{8 + 10} = \frac{-b}{-12 - 5} = \frac{c}{6 - 2}$
$\frac{a}{18} = \frac{b}{17} = \frac{c}{4} = k$
Thus,$a = 18k, b = 17k, c = 4k$.
Substituting these into $(i)$: $18k(x - 2) + 17k(y - 1) + 4k(z + 1) = 0$
$18x - 36 + 17y - 17 + 4z + 4 = 0$
$18x + 17y + 4z - 49 = 0$
Therefore,the equation is $18x + 17y + 4z = 49$.

(A) The equation of a plane passing through the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is given by $(x + 2y + 3z - 4) + \lambda(2x + y - z + 5) = 0$.
Rearranging the terms,we get: $x(1 + 2\lambda) + y(2 + \lambda) + z(3 - \lambda) + (5\lambda - 4) = 0 \quad \dots(i)$
Since this plane is perpendicular to the plane $5x + 3y + 6z + 8 = 0$,the dot product of their normal vectors must be zero $(a_1a_2 + b_1b_2 + c_1c_2 = 0)$:
$5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0$
$5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0$
$7\lambda + 29 = 0 \implies \lambda = -\frac{29}{7}$
Substituting $\lambda = -\frac{29}{7}$ into equation $(i)$:
$x(1 + 2(-\frac{29}{7})) + y(2 - \frac{29}{7}) + z(3 + \frac{29}{7}) + (5(-\frac{29}{7}) - 4) = 0$
$x(\frac{7 - 58}{7}) + y(\frac{14 - 29}{7}) + z(\frac{21 + 29}{7}) + (\frac{-145 - 28}{7}) = 0$
$-\frac{51}{7}x - \frac{15}{7}y + \frac{50}{7}z - \frac{173}{7} = 0$
Multiplying by $-7$,we get the required equation: $51x + 15y - 50z + 173 = 0$.

(A) The equation of the plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_{1} = d_{1}$ and $\vec{r} \cdot \vec{n}_{2} = d_{2}$ is given by $\vec{r} \cdot (\vec{n}_{1} + \lambda \vec{n}_{2}) = d_{1} + \lambda d_{2}$.
Given planes are $\vec{r} \cdot (\hat{i} + 3\hat{j}) = 6$ and $\vec{r} \cdot (3\hat{i} - \hat{j} - 4\hat{k}) = 0$.
So,the equation of the required plane is $\vec{r} \cdot [(\hat{i} + 3\hat{j}) + \lambda(3\hat{i} - \hat{j} - 4\hat{k})] = 6 + \lambda(0)$.
$\Rightarrow \vec{r} \cdot [(1 + 3\lambda)\hat{i} + (3 - \lambda)\hat{j} - 4\lambda\hat{k}] = 6$.
The perpendicular distance from the origin to the plane $\vec{r} \cdot \vec{n} = d$ is $\frac{|d|}{|\vec{n}|}$.
Here,$\frac{|6|}{\sqrt{(1 + 3\lambda)^{2} + (3 - \lambda)^{2} + (-4\lambda)^{2}}} = 1$.
$\Rightarrow 36 = (1 + 9\lambda^{2} + 6\lambda) + (9 + \lambda^{2} - 6\lambda) + 16\lambda^{2}$.
$\Rightarrow 36 = 26\lambda^{2} + 10$.
$\Rightarrow 26\lambda^{2} = 26 \Rightarrow \lambda^{2} = 1 \Rightarrow \lambda = \pm 1$.
For $\lambda = 1$,the equation is $\vec{r} \cdot (4\hat{i} + 2\hat{j} - 4\hat{k}) = 6 \Rightarrow 4x + 2y - 4z = 6 \Rightarrow 2x + y - 2z = 3$.
For $\lambda = -1$,the equation is $\vec{r} \cdot (-2\hat{i} + 4\hat{j} + 4\hat{k}) = 6 \Rightarrow -2x + 4y + 4z = 6 \Rightarrow -x + 2y + 2z = 3$.

(N/A) Let the given points be $\vec{a} = \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b} = 3\hat{i}+3\hat{j}+3\hat{k}$. The equation of the plane is $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$.
The distance $d$ of a point $\vec{p}$ from the plane $\vec{r} \cdot \vec{n} + d_0 = 0$ is given by $d = \frac{|\vec{p} \cdot \vec{n} + d_0|}{|\vec{n}|}$.
For point $\vec{a}$:
$d_1 = \frac{|(1)(5) + (-1)(2) + (3)(-7) + 9|}{\sqrt{5^2 + 2^2 + (-7)^2}} = \frac{|5 - 2 - 21 + 9|}{\sqrt{25 + 4 + 49}} = \frac{|-9|}{\sqrt{78}} = \frac{9}{\sqrt{78}}$.
For point $\vec{b}$:
$d_2 = \frac{|(3)(5) + (3)(2) + (3)(-7) + 9|}{\sqrt{5^2 + 2^2 + (-7)^2}} = \frac{|15 + 6 - 21 + 9|}{\sqrt{78}} = \frac{|9|}{\sqrt{78}} = \frac{9}{\sqrt{78}}$.
Since $d_1 = d_2$,the points are equidistant.
To check the sides,we evaluate $f(\vec{r}) = \vec{r} \cdot \vec{n} + d_0$ for both points:
$f(\vec{a}) = 5 - 2 - 21 + 9 = -9 < 0$.
$f(\vec{b}) = 15 + 6 - 21 + 9 = 9 > 0$.
Since the values have opposite signs,the points lie on opposite sides of the plane.

(N/A) Let the line $AB$ be represented by $\vec{r} = (6\hat{i} + 7\hat{j} + 4\hat{k}) + \lambda(3\hat{i} - \hat{j} + \hat{k})$. Any point $P$ on $AB$ is $(6 + 3\lambda, 7 - \lambda, 4 + \lambda)$.
Let the line $CD$ be represented by $\vec{r} = (0\hat{i} - 9\hat{j} + 2\hat{k}) + \mu(-3\hat{i} + 2\hat{j} + 4\hat{k})$. Any point $Q$ on $CD$ is $(-3\mu, -9 + 2\mu, 2 + 4\mu)$.
Then $\overrightarrow{PQ} = (-3\mu - 6 - 3\lambda)\hat{i} + (2\mu + \lambda - 16)\hat{j} + (4\mu - \lambda - 2)\hat{k}$.
Since $\overrightarrow{PQ} \perp \overrightarrow{AB}$,the dot product is $0$: $3(-3\mu - 6 - 3\lambda) - 1(2\mu + \lambda - 16) + 1(4\mu - \lambda - 2) = 0 \Rightarrow -7\mu - 11\lambda - 4 = 0 \dots (i)$.
Since $\overrightarrow{PQ} \perp \overrightarrow{CD}$,the dot product is $0$: $-3(-3\mu - 6 - 3\lambda) + 2(2\mu + \lambda - 16) + 4(4\mu - \lambda - 2) = 0 \Rightarrow 29\mu + 7\lambda - 22 = 0 \dots (ii)$.
Solving $(i)$ and $(ii)$,we multiply $(i)$ by $7$ and $(ii)$ by $11$: $-49\mu - 77\lambda - 28 = 0$ and $319\mu + 77\lambda - 242 = 0$.
Adding these,$270\mu - 270 = 0 \Rightarrow \mu = 1$. Substituting $\mu = 1$ in $(i)$,$-7 - 11\lambda - 4 = 0 \Rightarrow \lambda = -1$.
Thus,$P = (6 + 3(-1), 7 - (-1), 4 + (-1)) = (3, 8, 3)$ and $Q = (-3(1), -9 + 2(1), 2 + 4(1)) = (-3, -7, 6)$.
The position vectors are $\vec{P} = 3\hat{i} + 8\hat{j} + 3\hat{k}$ and $\vec{Q} = -3\hat{i} - 7\hat{j} + 6\hat{k}$.

(D) Since $C(0, -a, -1)$ lies on the plane $lx + my + nz = 0,$ we have $l(0) + m(-a) + n(-1) = 0,$ which implies $-ma - n = 0,$ or $\frac{m}{n} = -\frac{1}{a} \quad \dots(1)$
The vector $\vec{CA} = (a - 0, -2a - (-a), 3 - (-1)) = (a, -a, 4)$ is parallel to the normal vector of the plane $\vec{n} = (l, m, n).$
Thus,$\frac{a}{l} = \frac{-a}{m} = \frac{4}{n}.$ From $\frac{-a}{m} = \frac{4}{n},$ we get $\frac{m}{n} = -\frac{a}{4} \quad \dots(2)$
Equating $(1)$ and $(2),$ we have $-\frac{1}{a} = -\frac{a}{4} \Rightarrow a^2 = 4.$ Since $a > 0,$ we have $a = 2.$
Substituting $a = 2$ into $(1),$ we get $\frac{m}{n} = -\frac{1}{2}.$ Let $m = -t$ and $n = 2t.$ Then $\frac{2}{l} = \frac{-2}{-t} \Rightarrow l = t.$
The equation of the plane is $t(x - y + 2z) = 0,$ or $x - y + 2z = 0.$
The point $D$ is the foot of the perpendicular from $B(0, 4, 5)$ to the plane $x - y + 2z = 0.$ The line $BD$ is given by $\frac{x-0}{1} = \frac{y-4}{-1} = \frac{z-5}{2} = k.$
So,$D = (k, 4-k, 5+2k).$ Since $D$ lies on the plane,$k - (4-k) + 2(5+2k) = 0 \Rightarrow k - 4 + k + 10 + 4k = 0 \Rightarrow 6k + 6 = 0 \Rightarrow k = -1.$
Thus,$D = (-1, 4-(-1), 5+2(-1)) = (-1, 5, 3).$
With $C = (0, -2, -1)$ and $D = (-1, 5, 3),$ the length $CD = \sqrt{(-1-0)^2 + (5-(-2))^2 + (3-(-1))^2} = \sqrt{(-1)^2 + 7^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}.$

(C) The line is given by $\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3}$.
Since the plane $lx + my + nz = 0$ contains this line,it passes through the point $(1, -4, -2)$ and its normal vector $(l, m, n)$ is perpendicular to the direction vector $(-1, 2, 3)$ of the line.
Thus,$l(1) + m(-4) + n(-2) = 0 \implies l - 4m - 2n = 0$ (Equation $1$).
And $-l + 2m + 3n = 0$ (Equation $2$).
Adding $(1)$ and $(2)$,we get $-2m + n = 0 \implies n = 2m$.
Substituting $n = 2m$ into $(2)$,we get $-l + 2m + 3(2m) = 0 \implies l = 8m$.
Thus,the plane equation is $8x + y + 2z = 0$.
The point $C$ dividing $AB$ in ratio $k:1$ is $\left(\frac{2k-3}{k+1}, \frac{4k-6}{k+1}, \frac{-3k+1}{k+1}\right)$.
Since $C$ lies on the plane,$8\left(\frac{2k-3}{k+1}\right) + \left(\frac{4k-6}{k+1}\right) + 2\left(\frac{-3k+1}{k+1}\right) = 0$.
$16k - 24 + 4k - 6 - 6k + 2 = 0$.
$14k - 28 = 0 \implies k = 2$.

(B) The given line is $2x = 3y, z = 1$,which can be written as $\frac{x}{3} = \frac{y}{2}, z = 1$. The direction vector of this line is $\vec{v} = 3\hat{i} + 2\hat{j} + 0\hat{k}$.
The plane passes through points $A(1, 2, 1)$ and $B(2, 1, 2)$. The vector $\vec{AB} = (2-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = \hat{i} - \hat{j} + \hat{k}$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{AB}$ and $\vec{v}$:
$\vec{n} = \vec{AB} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 3 & 2 & 0 \end{vmatrix} = \hat{i}(0-2) - \hat{j}(0-3) + \hat{k}(2+3) = -2\hat{i} + 3\hat{j} + 5\hat{k}$.
The equation of the plane passing through $(1, 2, 1)$ with normal $\vec{n} = -2\hat{i} + 3\hat{j} + 5\hat{k}$ is:
$-2(x-1) + 3(y-2) + 5(z-1) = 0$
$-2x + 2 + 3y - 6 + 5z - 5 = 0$
$-2x + 3y + 5z - 9 = 0$ or $2x - 3y - 5z + 9 = 0$.
Now,check the options:
For $(-2, 0, 1)$: $2(-2) - 3(0) - 5(1) + 9 = -4 - 0 - 5 + 9 = 0$.
Thus,the plane passes through $(-2, 0, 1)$.

(C) The direction vectors of the two lines are $\vec{v}_1 = \hat{i} + \hat{j}$ and $\vec{v}_2 = \hat{j} - \hat{k}$.
The normal vector $\vec{n}$ to the plane $P$ is given by the cross product $\vec{v}_1 \times \vec{v}_2$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i}(-1) - \hat{j}(-1) + \hat{k}(1) = -\hat{i} + \hat{j} + \hat{k}$.
The plane passes through the point $(1, 0, 0)$ (from the first line). Thus,the equation of the plane is:
$-1(x - 1) + 1(y - 0) + 1(z - 0) = 0 \implies -x + y + z + 1 = 0 \implies x - y - z - 1 = 0$.
Let $Q(\alpha, \beta, \gamma)$ be the foot of the perpendicular from $M(1, 0, 1)$ to the plane $x - y - z - 1 = 0$. The line passing through $M$ and perpendicular to the plane is:
$\frac{\alpha - 1}{1} = \frac{\beta - 0}{-1} = \frac{\gamma - 1}{-1} = k$.
So,$\alpha = k + 1, \beta = -k, \gamma = 1 - k$.
Since $Q$ lies on the plane:
$(k + 1) - (-k) - (1 - k) - 1 = 0 \implies k + 1 + k - 1 + k - 1 = 0 \implies 3k - 1 = 0 \implies k = \frac{1}{3}$.
Thus,$\alpha = \frac{1}{3} + 1 = \frac{4}{3}, \beta = -\frac{1}{3}, \gamma = 1 - \frac{1}{3} = \frac{2}{3}$.
Finally,$3(\alpha + \beta + \gamma) = 3(\frac{4}{3} - \frac{1}{3} + \frac{2}{3}) = 3(\frac{5}{3}) = 5$.

(D) Let the points be $P(4,2,3)$,$A(1,-2,3)$,and $B(1,1,0)$.
The direction vector of line $AB$ is $\vec{v} = (1-1, 1-(-2), 0-3) = (0, 3, -3)$.
The equation of line $AB$ is $\vec{r} = (1, -2, 3) + \lambda(0, 3, -3) = (1, -2+3\lambda, 3-3\lambda)$.
Let $M$ be the foot of the perpendicular from $P$ to $AB$. Thus,$M = (1, -2+3\lambda, 3-3\lambda)$.
The vector $\vec{PM} = M - P = (1-4, -2+3\lambda-2, 3-3\lambda-3) = (-3, 3\lambda-4, -3\lambda)$.
Since $\vec{PM} \perp \vec{AB}$,their dot product is zero:
$(-3)(0) + (3\lambda-4)(3) + (-3\lambda)(-3) = 0$
$0 + 9\lambda - 12 + 9\lambda = 0$
$18\lambda = 12 \Rightarrow \lambda = \frac{2}{3}$.
Substituting $\lambda = \frac{2}{3}$ into $M$:
$M = (1, -2+3(\frac{2}{3}), 3-3(\frac{2}{3})) = (1, -2+2, 3-2) = (1, 0, 1)$.
Now,check which plane contains the point $(1, 0, 1)$:
For $2x+y-z=1$: $2(1) + 0 - 1 = 2 - 1 = 1$. This is correct.
Thus,the point $M$ lies on the plane $2x+y-z=1$.

(B) The equation of the line passing through $(1, -2, 3)$ and parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$ is given by $\frac{x-1}{2} = \frac{y+2}{3} = \frac{z-3}{-6} = r$.
Any point on this line is $(2r+1, 3r-2, -6r+3)$.
Since this point lies on the plane $x - y + z = 5$,we substitute these coordinates into the plane equation:
$(2r+1) - (3r-2) + (-6r+3) = 5$.
$2r + 1 - 3r + 2 - 6r + 3 = 5$.
$-7r + 6 = 5$.
$-7r = -1$.
$r = \frac{1}{7}$.
The distance between the point $(1, -2, 3)$ and the intersection point $(2r+1, 3r-2, -6r+3)$ is $\sqrt{(2r)^2 + (3r)^2 + (-6r)^2} = \sqrt{4r^2 + 9r^2 + 36r^2} = \sqrt{49r^2} = 7|r|$.
Substituting $r = \frac{1}{7}$,the distance is $7 \times \frac{1}{7} = 1$.