Plane Questions in English

(B) The normal vector $\vec{n}$ to the plane $P$ is the cross product of the direction vectors of the two lines:
$\vec{n} = (-2, 1, -3) \times (-1, 2, -2) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & -3 \\ -1 & 2 & -2 \end{vmatrix} = \hat{i}(-2 + 6) - \hat{j}(4 - 3) + \hat{k}(-4 + 1) = 4\hat{i} - \hat{j} - 3\hat{k}$.
The equation of the plane passing through $(2, 2, -2)$ with normal vector $(4, -1, -3)$ is:
$4(x - 2) - 1(y - 2) - 3(z + 2) = 0$
$4x - 8 - y + 2 - 3z - 6 = 0$
$4x - y - 3z = 12$.
To find the intercepts,divide by $12$:
$\frac{x}{3} + \frac{y}{-12} + \frac{z}{-4} = 1$.
Thus,$\alpha = 3, \beta = -12, \gamma = -4$.
Calculating $p = \alpha + \beta + \gamma = 3 - 12 - 4 = -13$.
The volume $V$ of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |\alpha \beta \gamma| = \frac{1}{6} |3 \times (-12) \times (-4)| = \frac{1}{6} |144| = 24$.
Therefore,the ordered pair $(V, p) = (24, -13)$.

(C) The equation of the plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant equation:
$\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}\right| = 0$
Substituting the points $(2, -3, 1)$,$(-1, 1, -2)$,and $(3, -4, 2)$:
$\left|\begin{array}{ccc} x-2 & y+3 & z-1 \\ -1-2 & 1-(-3) & -2-1 \\ 3-2 & -4-(-3) & 2-1 \end{array}\right| = 0$
$\left|\begin{array}{ccc} x-2 & y+3 & z-1 \\ -3 & 4 & -3 \\ 1 & -1 & 1 \end{array}\right| = 0$
Expanding the determinant:
$(x-2)(4 - 3) - (y+3)(-3 + 3) + (z-1)(3 - 4) = 0$
$(x-2)(1) - (y+3)(0) + (z-1)(-1) = 0$
$x - 2 - z + 1 = 0$
$x - z - 1 = 0$
The distance $d$ of a point $(x_0, y_0, z_0)$ from the plane $Ax + By + Cz + D = 0$ is $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
For the point $(7, -3, -4)$ and plane $x - z - 1 = 0$:
$d = \frac{|7 - (-4) - 1|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{|7 + 4 - 1|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5 \sqrt{2}$.

(C) Let the unit vector be $\hat{v} = \cos 60^{\circ} \hat{i} + \cos 45^{\circ} \hat{j} + \cos \gamma \hat{k}$.
Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,we have $\cos^2 60^{\circ} + \cos^2 45^{\circ} + \cos^2 \gamma = 1$.
$\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$.
$\Rightarrow \cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$.
Since $\gamma$ is an acute angle,$\cos \gamma = \frac{1}{2}$.
Thus,the normal vector to the plane is $\vec{n} = \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}$.
The equation of the plane passing through $(\sqrt{2}, -1, 1)$ is $\frac{1}{2}(x - \sqrt{2}) + \frac{1}{\sqrt{2}}(y + 1) + \frac{1}{2}(z - 1) = 0$.
Multiplying by $2$,we get $(x - \sqrt{2}) + \sqrt{2}(y + 1) + (z - 1) = 0$.
$\Rightarrow x - \sqrt{2} + \sqrt{2} y + \sqrt{2} + z - 1 = 0$.
$\Rightarrow x + \sqrt{2} y + z = 1$.
Since the point $(a, b, c)$ lies on the plane,we have $a + \sqrt{2} b + c = 1$.

(C) The direction vector of the line joining $(4, 1, -3)$ and $(2, 4, 3)$ is given by $\vec{n} = (2-4, 4-1, 3-(-3)) = (-2, 3, 6)$.
Since the plane $P$ is perpendicular to this line,the normal vector to the plane is $\vec{n} = (-2, 3, 6)$.
The equation of the plane passing through $(1, -1, -5)$ with normal vector $\vec{n} = (-2, 3, 6)$ is:
$-2(x - 1) + 3(y + 1) + 6(z + 5) = 0$
$-2x + 2 + 3y + 3 + 6z + 30 = 0$
$-2x + 3y + 6z + 35 = 0$ or $2x - 3y - 6z = 35$.
The distance of the point $(3, -2, 2)$ from the plane $2x - 3y - 6z - 35 = 0$ is given by the formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.
$d = \frac{|2(3) - 3(-2) - 6(2) - 35|}{\sqrt{2^2 + (-3)^2 + (-6)^2}}$
$d = \frac{|6 + 6 - 12 - 35|}{\sqrt{4 + 9 + 36}}$
$d = \frac{|-35|}{\sqrt{49}} = \frac{35}{7} = 5$.

(C) Let the plane $P$ be $2x + ay + 4z = k$. Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, a, 4)$,the normal vector to the plane is $\vec{n} = 2\hat{i} + a\hat{j} + 4\hat{k}$.
The equation of the plane is $2(x - 2) + a(y - a) + 4(z - 4) = 0$,which simplifies to $2x + ay + 4z = 4 + a^2 + 16 = 20 + a^2$.
The intercepts are $A = (\frac{20 + a^2}{2}, 0, 0)$,$B = (0, \frac{20 + a^2}{a}, 0)$,and $C = (0, 0, \frac{20 + a^2}{4})$.
The volume of the tetrahedron $OABC$ is $V = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} \cdot \frac{(20 + a^2)^3}{8a} = 144$.
$(20 + a^2)^3 = 144 \times 48 \times a = 6912a$.
Testing $a = 2$: $(20 + 4)^3 = 24^3 = 13824$ and $6912 \times 2 = 13824$. Thus,$a = 2$.
The equation of the plane is $2x + 2y + 4z = 20 + 4 = 24$,or $x + y + 2z = 12$.
Checking the points:
$A(2, 2, 4) \Rightarrow 2 + 2 + 2(4) = 12$ (Lies on plane).
$B(0, 4, 4) \Rightarrow 0 + 4 + 2(4) = 12$ (Lies on plane).
$C(3, 0, 4) \Rightarrow 3 + 0 + 2(4) = 11 \neq 12$ (Does $NOT$ lie on plane).
$D(0, 6, 3) \Rightarrow 0 + 6 + 2(3) = 12$ (Lies on plane).
Therefore,the point $(3, 0, 4)$ does not lie on the plane.

(B) The normal vector $\vec{n}$ of the plane is the direction vector of the line joining $(1, 2, 3)$ and $(-2, 3, 5)$.
$\vec{n} = (-2-1)\hat{i} + (3-2)\hat{j} + (5-3)\hat{k} = -3\hat{i} + \hat{j} + 2\hat{k}$.
Since the plane is perpendicular to this line,the normal vector is $\vec{n} = -3\hat{i} + \hat{j} + 2\hat{k}$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the point $(3, -2, 5)$ and normal vector $(-3, 1, 2)$:
$-3(x-3) + 1(y+2) + 2(z-5) = 0$.
$-3x + 9 + y + 2 + 2z - 10 = 0$.
$-3x + y + 2z = -1$.
Multiplying by $-1$ to match the form $\alpha x + \beta y + \gamma z = 1$:
$3x - y - 2z = 1$.
Comparing with $\alpha x + \beta y + \gamma z = 1$,we get $\alpha = 3$,$\beta = -1$,and $\gamma = -2$.
The product $\alpha \beta \gamma = (3)(-1)(-2) = 6$.

(A) The equation of a plane passing through three points $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by the normal vector $\vec{n} = \vec{AB} \times \vec{AC}$.
$\vec{AB} = (1-1, 4-2, 1-0) = (0, 2, 1)$
$\vec{AC} = (0-1, 5-2, 1-0) = (-1, 3, 1)$
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 1 \\ -1 & 3 & 1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(0+1) + \hat{k}(0+2) = -\hat{i} - \hat{j} + 2\hat{k}$.
The equation of the plane is $-1(x-1) - 1(y-2) + 2(z-0) = 0$,which simplifies to $x + y - 2z - 3 = 0$.
Let the image of $P(1, 2, 6)$ be $Q(\alpha, \beta, \gamma)$. The formula for the image is $\frac{\alpha - x_0}{a} = \frac{\beta - y_0}{b} = \frac{\gamma - z_0}{c} = -2 \frac{ax_0 + by_0 + cz_0 + d}{a^2 + b^2 + c^2}$.
Substituting the values: $\frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = \frac{\gamma - 6}{-2} = -2 \frac{1(1) + 1(2) - 2(6) - 3}{1^2 + 1^2 + (-2)^2} = -2 \frac{1 + 2 - 12 - 3}{6} = -2 \frac{-12}{6} = 4$.
Thus,$\alpha - 1 = 4 \Rightarrow \alpha = 5$,$\beta - 2 = 4 \Rightarrow \beta = 6$,and $\gamma - 6 = -8 \Rightarrow \gamma = -2$.
Finally,$\alpha^2 + \beta^2 + \gamma^2 = 5^2 + 6^2 + (-2)^2 = 25 + 36 + 4 = 65$.

(D) The equation of a plane passing through the point $(-2, 3, 5)$ is given by $a(x + 2) + b(y - 3) + c(z - 5) = 0$.
Since this plane is perpendicular to the planes $2x + 4y + 5z = 8$ and $3x - 2y + 3z = 5$,its normal vector $(a, b, c)$ must be perpendicular to the normal vectors of the given planes,which are $\vec{n_1} = (2, 4, 5)$ and $\vec{n_2} = (3, -2, 3)$.
Thus,the normal vector $(a, b, c)$ is proportional to the cross product $\vec{n_1} \times \vec{n_2}$:
$(a, b, c) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 5 \\ 3 & -2 & 3 \end{vmatrix} = \hat{i}(12 - (-10)) - \hat{j}(6 - 15) + \hat{k}(-4 - 12) = 22\hat{i} + 9\hat{j} - 16\hat{k}$.
So,$a = 22, b = 9, c = -16$.
The equation of the plane is $22(x + 2) + 9(y - 3) - 16(z - 5) = 0$.
Expanding this,we get $22x + 44 + 9y - 27 - 16z + 80 = 0$,which simplifies to $22x + 9y - 16z + 97 = 0$.
Comparing this with $\alpha x + \beta y + \gamma z + 97 = 0$,we find $\alpha = 22, \beta = 9, \gamma = -16$.
Therefore,$\alpha + \beta + \gamma = 22 + 9 - 16 = 15$.

(A) The formula for the image $(\alpha, \beta, \gamma)$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is given by:
$\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = -2 \left( \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2} \right)$
Given point $P(2, 3, 5)$ and plane $2x + y - 3z - 6 = 0$,we have $a=2, b=1, c=-3, d=-6$:
$\frac{\alpha - 2}{2} = \frac{\beta - 3}{1} = \frac{\gamma - 5}{-3} = -2 \left( \frac{2(2) + 1(3) - 3(5) - 6}{2^2 + 1^2 + (-3)^2} \right)$
Calculate the value inside the bracket:
$\frac{4 + 3 - 15 - 6}{4 + 1 + 9} = \frac{-14}{14} = -1$
So,the ratio is $-2(-1) = 2$:
$\frac{\alpha - 2}{2} = 2 \implies \alpha - 2 = 4 \implies \alpha = 6$
$\frac{\beta - 3}{1} = 2 \implies \beta - 3 = 2 \implies \beta = 5$
$\frac{\gamma - 5}{-3} = 2 \implies \gamma - 5 = -6 \implies \gamma = -1$
Therefore,$\alpha + \beta + \gamma = 6 + 5 - 1 = 10$.

(B) The plane $P$ passes through points $Q(5,3,0)$,$R(13,3,-2)$,and $S(1,6,2)$.
Vectors in the plane are $\vec{QR} = (13-5, 3-3, -2-0) = (8, 0, -2)$ and $\vec{QS} = (1-5, 6-3, 2-0) = (-4, 3, 2)$.
The normal vector $\vec{n} = \vec{QR} \times \vec{QS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & 0 & -2 \\ -4 & 3 & 2 \end{vmatrix} = \hat{i}(0 - (-6)) - \hat{j}(16 - 8) + \hat{k}(24 - 0) = 6\hat{i} - 8\hat{j} + 24\hat{k}$.
Dividing by $2$,we get the normal vector $\vec{n}' = 3\hat{i} - 4\hat{j} + 12\hat{k}$.
The equation of the plane is $3(x-5) - 4(y-3) + 12(z-0) = 0$,which simplifies to $3x - 4y + 12z = 3$.
The distance of point $A(3,4,\alpha)$ from the plane is $\frac{|3(3) - 4(4) + 12(\alpha) - 3|}{\sqrt{3^2 + (-4)^2 + 12^2}} = \frac{|9 - 16 + 12\alpha - 3|}{13} = \frac{|12\alpha - 10|}{13} = 2$.
$|12\alpha - 10| = 26 \implies 12\alpha - 10 = 26$ or $12\alpha - 10 = -26$.
$12\alpha = 36 \implies \alpha = 3$ (since $\alpha \in N$).
Now,the distance of point $B(2,3,a)$ from the plane is $\frac{|3(2) - 4(3) + 12(a) - 3|}{13} = 3$.
$|6 - 12 + 12a - 3| = 39 \implies |12a - 9| = 39$.
$12a - 9 = 39 \implies 12a = 48 \implies a = 4$.
$12a - 9 = -39 \implies 12a = -30 \implies a = -2.5$.
Since we need the positive value,$a = 4$.

(D) The plane equation is $x+3y-2z+6=0$. Setting two coordinates to zero,we find the intercepts:
$A(-6, 0, 0)$,$B(0, -2, 0)$,$C(0, 0, 3)$.
Let $H(\alpha, \beta, \frac{6}{7})$ be the orthocentre.
Since $H$ lies on the plane $ABC$,$\alpha + 3\beta - 2(\frac{6}{7}) + 6 = 0 \implies \alpha + 3\beta = -6 + \frac{12}{7} = -\frac{30}{7}$.
Also,$\overrightarrow{AH} \cdot \overrightarrow{BC} = 0$.
$\overrightarrow{AH} = (\alpha+6, \beta, \frac{6}{7}-0) = (\alpha+6, \beta, \frac{6}{7})$.
$\overrightarrow{BC} = (0, 2, 3)$.
$(\alpha+6)(0) + \beta(2) + \frac{6}{7}(3) = 0 \implies 2\beta + \frac{18}{7} = 0 \implies \beta = -\frac{9}{7}$.
Substituting $\beta$ into the plane equation: $\alpha + 3(-\frac{9}{7}) = -\frac{30}{7} \implies \alpha - \frac{27}{7} = -\frac{30}{7} \implies \alpha = -\frac{3}{7}$.
Now,$98(\alpha+\beta)^2 = 98(-\frac{3}{7} - \frac{9}{7})^2 = 98(-\frac{12}{7})^2 = 98 \times \frac{144}{49} = 2 \times 144 = 288$.

(C) Let the two points be $A(0,-1,2)$ and $B(-1,2,1)$. The vector $\vec{AB} = (-1-0)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$.
The line is parallel to the vector $\vec{v}$ passing through $P(5,1,-7)$ and $Q(1,-1,-1)$,so $\vec{v} = (1-5)\hat{i} + (-1-1)\hat{j} + (-1-(-7))\hat{k} = -4\hat{i} - 2\hat{j} + 6\hat{k}$.
The normal vector $\vec{n}$ to the plane is $\vec{AB} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & -1 \\ -4 & -2 & 6 \end{vmatrix} = \hat{i}(18-2) - \hat{j}(-6-4) + \hat{k}(2+12) = 16\hat{i} + 10\hat{j} + 14\hat{k}$.
Dividing by $2$,we get the normal vector $\vec{n}' = 8\hat{i} + 5\hat{j} + 7\hat{k}$.
The equation of the plane is $8x + 5y + 7z = d$. Substituting point $(0,-1,2)$,we get $8(0) + 5(-1) + 7(2) = -5 + 14 = 9$. So,$d = 9$.
The equation of the plane is $8x + 5y + 7z = 9$.
Checking the options: For $(0,5,-2)$,$8(0) + 5(5) + 7(-2) = 25 - 14 = 11 \neq 9$. For $(-2,5,0)$,$8(-2) + 5(5) + 7(0) = -16 + 25 = 9$. Thus,the plane passes through $(-2,5,0)$.

(D) The normal vectors to the planes are $\vec{n}_1 = (1, -1, 1)$,$\vec{n}_2 = (1, 1, -1)$,and $\vec{n}_3 = (1, -3, 3)$.
The direction vector of line $L_1$ (intersection of $P_2$ and $P_3$) is $\vec{v}_1 = \vec{n}_2 \times \vec{n}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -3 & 3 \end{vmatrix} = (0, -4, -4)$,which is parallel to $(0, 1, 1)$.
The direction vector of line $L_2$ (intersection of $P_3$ and $P_1$) is $\vec{v}_2 = \vec{n}_3 \times \vec{n}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 3 \\ 1 & -1 & 1 \end{vmatrix} = (0, 2, 2)$,which is parallel to $(0, 1, 1)$.
The direction vector of line $L_3$ (intersection of $P_1$ and $P_2$) is $\vec{v}_3 = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = (0, 2, 2)$,which is parallel to $(0, 1, 1)$.
Since all lines $L_1, L_2, L_3$ have the same direction vector $(0, 1, 1)$,they are all parallel to each other. Thus,$STATEMENT-1$ is False.
For $STATEMENT-2$,we check if the system has a common point by solving the equations. Adding $P_1$ and $P_2$ gives $2x = 0$,so $x = 0$. Substituting $x=0$ into $P_1$ and $P_2$ gives $-y+z=1$ and $y-z=-1$,which are identical. Substituting $x=0$ into $P_3$ gives $-3y+3z=2$,or $-y+z=2/3$. Since $1 \neq 2/3$,the system has no common point. Thus,$STATEMENT-2$ is True.

(C) Let the points be $P(1, 0, -1)$ and $Q(3, 2, -1)$. The mirror image of $P$ with respect to the plane is $Q$.
The midpoint $R$ of the line segment $PQ$ lies on the plane.
$R = \left( \frac{1+3}{2}, \frac{0+2}{2}, \frac{-1-1}{2} \right) = (2, 1, -1)$.
Since $R$ lies on the plane $\alpha x + \beta y + \gamma z = \delta$,we have:
$2\alpha + \beta - \gamma = \delta$ --- $(1)$
The normal vector to the plane is $\vec{n} = (\alpha, \beta, \gamma)$.
The vector $\vec{PQ} = (3-1, 2-0, -1-(-1)) = (2, 2, 0)$ is parallel to the normal vector $\vec{n}$.
Thus,$\frac{\alpha}{2} = \frac{\beta}{2} = \frac{\gamma}{0} = k$ (where $k \neq 0$).
This gives $\alpha = 2k$,$\beta = 2k$,and $\gamma = 0$.
Given $\alpha + \gamma = 1$,we have $2k + 0 = 1$,so $k = \frac{1}{2}$.
Therefore,$\alpha = 1$,$\beta = 1$,and $\gamma = 0$.
Substituting these into $(1)$:
$2(1) + 1(1) - 0 = \delta \implies \delta = 3$.
Now,check the options:
$(A)$ $\alpha + \beta = 1 + 1 = 2$ (True)
$(B)$ $\delta - \gamma = 3 - 0 = 3$ (True)
$(C)$ $\delta + \beta = 3 + 1 = 4$ (True)
$(D)$ $\alpha + \beta + \gamma = 1 + 1 + 0 = 2 \neq \delta$ (False)
Thus,statements $A, B, C$ are true.

(B) Let $X = (x, y, z)$. The condition for $S$ is $(x-1)^2 + (y-2)^2 + (z-3)^2 - ((x-4)^2 + (y-2)^2 + (z-7)^2) = 50$.
Expanding this,we get $(x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9) - (x^2 - 8x + 16 + y^2 - 4y + 4 + z^2 - 14z + 49) = 50$.
Simplifying,$(6x + 8z - 50) = 50$,so $6x + 8z = 100$,or $3x + 4z = 50$. This is a plane.
Similarly,for $T$,we get $(x-4)^2 + (y-2)^2 + (z-7)^2 - ((x-1)^2 + (y-2)^2 + (z-3)^2) = 50$.
This simplifies to $-(6x + 8z - 50) = 50$,so $6x + 8z = 0$,or $3x + 4z = 0$. This is a parallel plane.
$(A)$ Since $S$ is a plane,we can always find three non-collinear points in it to form a triangle of any area,including $1$. Thus,$(A)$ is $TRUE$.
$(B)$ Since $T$ is a plane,any line segment connecting two points $L, M \in T$ lies entirely within the plane $T$. Thus,$(B)$ is $TRUE$.
$(C)$ The distance between the parallel planes $3x + 4z - 50 = 0$ and $3x + 4z = 0$ is $d = \frac{|50 - 0|}{\sqrt{3^2 + 4^2}} = \frac{50}{5} = 10$.
For a rectangle with two vertices in $S$ and two in $T$,let the side length in the plane be $a$ and the side length connecting the planes be $b$. The distance between planes is $10$,so $b = 10$. Perimeter $2(a + 10) = 48 \implies a + 10 = 24 \implies a = 14$. Since we can choose any line of length $14$ in $S$,there are infinitely many such rectangles. Thus,$(C)$ is $TRUE$.
$(D)$ For a square,$a = b = 10$. Perimeter $4a = 40 \neq 48$. However,the question asks if there is a square of perimeter $48$,which would require $a = 12$. Since the distance between planes is $10$,we can form a rectangle with sides $12$ and $10$. If we need a square,we need $a = 10$,so perimeter $40$. The statement $(D)$ claims a square of perimeter $48$ exists,which is impossible as $a$ must be $10$. Wait,checking the logic: if $a=12$ and $b=10$,it is a rectangle,not a square. Thus $(D)$ is $FALSE$.

(C) The equation of the plane is $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$.
To find the points where the plane cuts the coordinate axes,we set the other two variables to zero.
For the $x$-axis,set $y=0$ and $z=0$: $\frac{x}{3}=1 \implies x=3$. So,$A = (3, 0, 0)$.
For the $y$-axis,set $x=0$ and $z=0$: $\frac{y}{2}=1 \implies y=2$. So,$B = (0, 2, 0)$.
For the $z$-axis,set $x=0$ and $y=0$: $-\frac{z}{4}=1 \implies z=-4$. So,$C = (0, 0, -4)$.
The area of a triangle with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.
$\vec{AB} = B - A = (-3, 2, 0)$ and $\vec{AC} = C - A = (-3, 0, -4)$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 0 \\ -3 & 0 & -4 \end{vmatrix} = \hat{i}(-8 - 0) - \hat{j}(12 - 0) + \hat{k}(0 - (-6)) = -8\hat{i} - 12\hat{j} + 6\hat{k}$.
The magnitude is $\sqrt{(-8)^2 + (-12)^2 + 6^2} = \sqrt{64 + 144 + 36} = \sqrt{244} = \sqrt{4 \times 61} = 2\sqrt{61}$.
Area $= \frac{1}{2} \times 2\sqrt{61} = \sqrt{61}$ sq. units.

(A) Let $A = (1, 2, 3)$ and $B = \left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. The plane passes through the midpoint $M$ of $AB$ and is perpendicular to the line segment $AB$.
The midpoint $M$ is given by:
$M = \left( \frac{1 - \frac{7}{3}}{2}, \frac{2 - \frac{4}{3}}{2}, \frac{3 - \frac{1}{3}}{2} \right) = \left( \frac{-\frac{4}{3}}{2}, \frac{\frac{2}{3}}{2}, \frac{\frac{8}{3}}{2} \right) = \left( -\frac{2}{3}, \frac{1}{3}, \frac{4}{3} \right)$.
The direction ratios of the normal to the plane are the same as the direction ratios of the line $AB$:
$D.r.s = \left( 1 - (-\frac{7}{3}), 2 - (-\frac{4}{3}), 3 - (-\frac{1}{3}) \right) = \left( \frac{10}{3}, \frac{10}{3}, \frac{10}{3} \right)$.
We can take the normal vector as $\vec{n} = (1, 1, 1)$.
The equation of the plane is $1(x - x_0) + 1(y - y_0) + 1(z - z_0) = 0$ where $(x_0, y_0, z_0) = M = \left( -\frac{2}{3}, \frac{1}{3}, \frac{4}{3} \right)$:
$1(x + \frac{2}{3}) + 1(y - \frac{1}{3}) + 1(z - \frac{4}{3}) = 0$
$x + y + z + \frac{2}{3} - \frac{1}{3} - \frac{4}{3} = 0$
$x + y + z - 1 = 0 \Rightarrow x + y + z = 1$.
Checking the options:
$(A)$ $(1, -1, 1) \Rightarrow 1 - 1 + 1 = 1$. This point lies on the plane.

(D) The normal vector to face $OPQ$ is given by the cross product of vectors $\vec{OP}$ and $\vec{OQ}$.
$\vec{n_1} = \vec{OP} \times \vec{OQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ -1 & 1 & 2 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(4+3) + \hat{k}(2+1) = -\hat{i} - 7\hat{j} + 3\hat{k}$.
The normal vector to face $PQR$ is given by the cross product of vectors $\vec{PQ}$ and $\vec{PR}$.
$\vec{PQ} = (-1-2, 1-1, 2-3) = (-3, 0, -1)$ and $\vec{PR} = (1-2, 2-1, 1-3) = (-1, 1, -2)$.
$\vec{n_2} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & -1 \\ -1 & 1 & -2 \end{vmatrix} = \hat{i}(0+1) - \hat{j}(6-1) + \hat{k}(-3-0) = \hat{i} - 5\hat{j} - 3\hat{k}$.
The angle $\theta$ between the faces is the angle between their normal vectors:
$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|(-1)(1) + (-7)(-5) + (3)(-3)|}{\sqrt{(-1)^2+(-7)^2+3^2} \sqrt{1^2+(-5)^2+(-3)^2}}$
$\cos \theta = \frac{|-1 + 35 - 9|}{\sqrt{1+49+9} \sqrt{1+25+9}} = \frac{25}{\sqrt{59} \sqrt{35}}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{25}{\sqrt{59} \sqrt{35}}\right)$.

(D) The given equation of the plane is in the form $\bar{r} = \bar{a} + \lambda \bar{A} + \mu \bar{B}$,where $\bar{a} = \hat{i} - \hat{j}$,$\bar{A} = \hat{i} + \hat{j} + \hat{k}$,and $\bar{B} = \hat{i} - 2\hat{j} + 3\hat{k}$.
The normal vector $\bar{n}$ to the plane is given by the cross product of the two vectors $\bar{A}$ and $\bar{B}$:
$\bar{n} = \bar{A} \times \bar{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix}$
$= \hat{i}(3 - (-2)) - \hat{j}(3 - 1) + \hat{k}(-2 - 1)$
$= 5\hat{i} - 2\hat{j} - 3\hat{k}$.
The Cartesian equation of the plane is given by $(\bar{r} - \bar{a}) \cdot \bar{n} = 0$,which is $\bar{r} \cdot \bar{n} = \bar{a} \cdot \bar{n}$.
Calculating $\bar{a} \cdot \bar{n} = (\hat{i} - \hat{j}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = (1)(5) + (-1)(-2) + (0)(-3) = 5 + 2 = 7$.
Thus,the Cartesian equation is $5x - 2y - 3z = 7$,or $5x - 2y - 3z - 7 = 0$.

(C) Two planes $\bar{r} \cdot \bar{n}_1 = d_1$ and $\bar{r} \cdot \bar{n}_2 = d_2$ are parallel if their normal vectors $\bar{n}_1$ and $\bar{n}_2$ are proportional.
Here,$\bar{n}_1 = 2 \hat{i} - \lambda \hat{j} + \hat{k}$ and $\bar{n}_2 = 4 \hat{i} - \hat{j} + \mu \hat{k}$.
Since they are parallel,we have $\frac{2}{4} = \frac{-\lambda}{-1} = \frac{1}{\mu}$.
From $\frac{2}{4} = \frac{\lambda}{1}$,we get $\lambda = \frac{1}{2}$.
From $\frac{2}{4} = \frac{1}{\mu}$,we get $\mu = 2$.
Therefore,$\lambda + \mu = \frac{1}{2} + 2 = \frac{5}{2}$.

(C) Let the points be $A(4,2,3)$,$B(-1,4,2)$,and $C(3,2,1)$.
Two vectors lying on the plane are $\vec{AB} = (-1-4)\hat{i} + (4-2)\hat{j} + (2-3)\hat{k} = -5\hat{i} + 2\hat{j} - 1\hat{k}$ and $\vec{AC} = (3-4)\hat{i} + (2-2)\hat{j} + (1-3)\hat{k} = -1\hat{i} + 0\hat{j} - 2\hat{k}$.
The normal vector $\vec{n}$ is given by $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 2 & -1 \\ -1 & 0 & -2 \end{vmatrix} = \hat{i}(-4-0) - \hat{j}(10-1) + \hat{k}(0 - (-2)) = -4\hat{i} - 9\hat{j} + 2\hat{k}$.
The magnitude of $\vec{n}$ is $|\vec{n}| = \sqrt{(-4)^2 + (-9)^2 + 2^2} = \sqrt{16 + 81 + 4} = \sqrt{101}$.
The direction cosines are $\frac{-4}{\sqrt{101}}, \frac{-9}{\sqrt{101}}, \frac{2}{\sqrt{101}}$.

(D) Two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are parallel if their normal vectors are proportional,i.e.,$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given planes are $2x - 5y + z - 8 = 0$ and $2\lambda x - 15y + \lambda z + 6 = 0$.
Comparing the coefficients,we get:
$\frac{2}{2\lambda} = \frac{-5}{-15} = \frac{1}{\lambda}$.
Simplifying the ratios:
$\frac{1}{\lambda} = \frac{1}{3} = \frac{1}{\lambda}$.
From this,we conclude that $\lambda = 3$.

(D) The equation of the family of planes passing through the line of intersection of the planes $x+2y+3z-4=0$ and $4x+3y+2z-1=0$ is given by $(x+2y+3z-4) + \lambda(4x+3y+2z-1) = 0$.
Since the plane passes through the origin $(0, 0, 0)$,we substitute $x=0, y=0, z=0$ into the equation:
$(0+0+0-4) + \lambda(0+0+0-1) = 0
\Rightarrow -4 - \lambda = 0
\Rightarrow \lambda = -4$.
Substituting $\lambda = -4$ back into the family equation:
$(x+2y+3z-4) - 4(4x+3y+2z-1) = 0
\Rightarrow x+2y+3z-4 - 16x - 12y - 8z + 4 = 0
\Rightarrow -15x - 10y - 5z = 0
\Rightarrow 3x + 2y + z = 0$.
The direction ratios of the normal to the plane $ax+by+cz+d=0$ are $(a, b, c)$.
Thus,the direction ratios of the normal to the plane $3x+2y+z=0$ are $3, 2, 1$.

(C) The given equation of the plane is in the form $\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}$,where $\vec{a} = 2 \hat{i} - 3 \hat{j}$,$\vec{b} = \hat{i} + 2 \hat{j} - \hat{k}$,and $\vec{c} = 2 \hat{i} + 3 \hat{j} + \hat{k}$.
To find the Cartesian equation,we need the normal vector $\vec{n} = \vec{b} \times \vec{c}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(2+3) - \hat{j}(1+2) + \hat{k}(3-4) = 5 \hat{i} - 3 \hat{j} - \hat{k}$.
The Cartesian equation of the plane is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which is $(x-2, y+3, z-0) \cdot (5, -3, -1) = 0$.
$5(x-2) - 3(y+3) - 1(z) = 0$.
$5x - 10 - 3y - 9 - z = 0$.
$5x - 3y - z = 19$.

(C) The normal vector $\vec{n}$ to the plane is along $\overrightarrow{AB}$.
$\overrightarrow{AB} = \vec{B} - \vec{A} = (\hat{i}-2 \hat{j}-4 \hat{k}) - (3 \hat{i}+\hat{j}+2 \hat{k}) = -2 \hat{i}-3 \hat{j}-6 \hat{k}$.
Since the plane is perpendicular to $\overrightarrow{AB}$,the normal vector to the plane can be taken as $\vec{n} = 2 \hat{i}+3 \hat{j}+6 \hat{k}$.
The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = a \hat{i}+b \hat{j}+c \hat{k}$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the point $B(1, -2, -4)$ and the normal vector components $(2, 3, 6)$:
$2(x-1) + 3(y+2) + 6(z+4) = 0$.
$2x - 2 + 3y + 6 + 6z + 24 = 0$.
$2x + 3y + 6z + 28 = 0$.

(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of the given planes $\vec{n}_1 = 2\hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{n}_2 = \hat{i} + 2\hat{j} - 3\hat{k}$.
Thus,$\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -2 \\ 1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 + 4) - \hat{j}(-6 + 2) + \hat{k}(4 - 3) = -5\hat{i} + 4\hat{j} + \hat{k}$.
The equation of the plane passing through $(x_0, y_0, z_0) = (1, -1, 2)$ with normal $\vec{n} = -5\hat{i} + 4\hat{j} + \hat{k}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
$\vec{a} \cdot \vec{n} = (1\hat{i} - 1\hat{j} + 2\hat{k}) \cdot (-5\hat{i} + 4\hat{j} + \hat{k}) = (1)(-5) + (-1)(4) + (2)(1) = -5 - 4 + 2 = -7$.
Therefore,the vector equation is $\vec{r} \cdot (-5\hat{i} + 4\hat{j} + \hat{k}) = -7$.

(A) The normal vector $\vec{n}$ to the plane is the vector from the origin $O(0, 0, 0)$ to the foot of the perpendicular $M(-1, -2, 2)$.
Thus,$\vec{n} = \vec{OM} = -\hat{i} - 2\hat{j} + 2\hat{k}$.
The equation of a plane passing through a point $M$ with normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{OM} \cdot \vec{n}$.
Here,$\vec{OM} \cdot \vec{n} = (-1, -2, 2) \cdot (-1, -2, 2) = (-1)^2 + (-2)^2 + (2)^2 = 1 + 4 + 4 = 9$.
Therefore,the vector equation of the plane is $\vec{r} \cdot(-\hat{i} - 2\hat{j} + 2\hat{k}) = 9$.

(C) Let the equation of the plane be $a(x - 1) + b(y + 2) + c(z - 1) = 0$,where $\vec{n} = (a, b, c)$ is the normal vector to the plane.
Since the plane is perpendicular to $2x - 2y + z = 0$ and $x - y + 2z = 4$,the normal vector $\vec{n}$ is parallel to the cross product of the normals of the given planes,$\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$.
$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.
We can take the normal vector as $\vec{n} = (1, 1, 0)$.
The equation of the plane is $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$,which simplifies to $x + y + 1 = 0$.
The distance of the point $(1, 2, 2)$ from the plane $x + y + 1 = 0$ is given by $d = \frac{|1 + 2 + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ units.

(C) The equation of the $XY$-plane is $z = 0$.
Any plane parallel to the $XY$-plane is of the form $z = k$,where $k$ is a constant.
Since the plane passes through the point $A(7, 8, 6)$,the $z$-coordinate of the plane must be equal to the $z$-coordinate of the point $A$.
Therefore,$k = 6$.
Thus,the equation of the plane is $z = 6$.

(C) The plane passes through $A(2,1,2)$ and $B(1,2,1)$. The vector $\vec{AB} = (1-2, 2-1, 1-2) = (-1, 1, -1)$.
The line is given by $2x = 3y$ and $z = 1$. This can be written as $\frac{x}{3} = \frac{y}{2}$ and $z = 1$. The direction vector of the line is $\vec{v} = (3, 2, 0)$.
The normal vector $\vec{n}$ to the plane is $\vec{AB} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -1 \\ 3 & 2 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-3)) + \hat{k}(-2 - 3) = 2\hat{i} - 3\hat{j} - 5\hat{k}$.
The equation of the plane is $2(x-2) - 3(y-1) - 5(z-2) = 0$,which simplifies to $2x - 4 - 3y + 3 - 5z + 10 = 0$,or $2x - 3y - 5z + 9 = 0$.
Checking the options:
For $(-6, 2, 0)$: $2(-6) - 3(2) - 5(0) + 9 = -12 - 6 + 9 = -9 \neq 0$.
For $(6, -2, 0)$: $2(6) - 3(-2) - 5(0) + 9 = 12 + 6 + 9 = 27 \neq 0$.
For $(-2, 0, 1)$: $2(-2) - 3(0) - 5(1) + 9 = -4 - 5 + 9 = 0$. Thus,the plane passes through $(-2, 0, 1)$.

(B) The normal vectors to the planes are $\vec{n_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{n_2} = \hat{i} + \alpha\hat{j} + 2\hat{k}$.
The angle $\theta$ between the planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Given $\cos \theta = \frac{1}{14}$,we have $\frac{|(1)(1) + (-2)(\alpha) + (3)(2)|}{\sqrt{1^2 + (-2)^2 + 3^2} \sqrt{1^2 + \alpha^2 + 2^2}} = \frac{1}{14}$.
$\frac{|7 - 2\alpha|}{\sqrt{14} \sqrt{\alpha^2 + 5}} = \frac{1}{14}$.
Squaring both sides: $\frac{(7 - 2\alpha)^2}{14(\alpha^2 + 5)} = \frac{1}{196}$.
$\frac{49 - 28\alpha + 4\alpha^2}{\alpha^2 + 5} = \frac{14}{196} = \frac{1}{14}$.
$14(4\alpha^2 - 28\alpha + 49) = \alpha^2 + 5$.
$56\alpha^2 - 392\alpha + 686 = \alpha^2 + 5$.
$55\alpha^2 - 392\alpha + 681 = 0$.
Let the roots be $\alpha_1$ and $\alpha_2$. The difference is $|\alpha_1 - \alpha_2| = \frac{\sqrt{D}}{|a|} = \frac{\sqrt{(-392)^2 - 4(55)(681)}}{55}$.
$D = 153664 - 149820 = 3844$.
$\sqrt{D} = \sqrt{3844} = 62$.
Difference $= \frac{62}{55}$.

(A) The equation of the plane is given in parametric form as $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$,where $\vec{a} = \hat{i} - \hat{j}$,$\vec{b} = \hat{i} + \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} - 2\hat{j} + 3\hat{k}$.
To find the normal vector $\vec{n}$ to the plane,we calculate the cross product $\vec{n} = \vec{b} \times \vec{c}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i}(3 - (-2)) - \hat{j}(3 - 1) + \hat{k}(-2 - 1) = 5\hat{i} - 2\hat{j} - 3\hat{k}$.
The equation of the plane in scalar form is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Calculating $\vec{a} \cdot \vec{n} = (\hat{i} - \hat{j}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = (1)(5) + (-1)(-2) + (0)(-3) = 5 + 2 = 7$.
So,the equation of the plane is $5x - 2y - 3z = 7$.
The distance $d$ of the plane $Ax + By + Cz = D$ from the origin $(0, 0, 0)$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 5, B = -2, C = -3$,and $D = 7$.
$d = \frac{|7|}{\sqrt{5^2 + (-2)^2 + (-3)^2}} = \frac{7}{\sqrt{25 + 4 + 9}} = \frac{7}{\sqrt{38}}$ units.

(A) The line passes through $A(1,3,2)$ and $B(1,2,1)$. The direction vector of the line is $\vec{v} = B - A = (1-1, 2-3, 1-2) = (0, -1, -1)$.
The plane passes through $P(2,1,-1)$ and $A(1,3,2)$. The vector $\vec{PA} = (1-2, 3-1, 2-(-1)) = (-1, 2, 3)$.
The normal to the plane is $\vec{n} = \vec{v} \times \vec{PA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -1 & -1 \\ -1 & 2 & 3 \end{vmatrix} = \hat{i}(-3+2) - \hat{j}(0-1) + \hat{k}(0-1) = -\hat{i} + \hat{j} - \hat{k}$.
The equation of the plane is $-1(x-2) + 1(y-1) - 1(z+1) = 0$,which simplifies to $-x+2+y-1-z-1 = 0$,or $-x+y-z = 0$,i.e.,$x-y+z = 0$.
Since the plane passes through the origin $(0,0,0)$,the intercepts $p, q, r$ are all $0$.
Therefore,$p+q+r = 0+0+0 = 0$.

(A) The equation of a plane passing through a point $P(x_1, y_1, z_1)$ and having a normal vector $\vec{n} = (a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Here,the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $P(2, -1, 4)$.
This means the vector $\vec{OP} = (2, -1, 4)$ is the normal vector to the plane.
So,$a = 2, b = -1, c = 4$.
The equation of the plane is $2(x-2) - 1(y-(-1)) + 4(z-4) = 0$.
$2(x-2) - 1(y+1) + 4(z-4) = 0$.
$2x - 4 - y - 1 + 4z - 16 = 0$.
$2x - y + 4z - 21 = 0$.

(C) The given equation is $x^2 - 8x + 12 = 0$.
Factoring the quadratic equation,we get $(x - 6)(x - 2) = 0$.
This implies $x = 6$ or $x = 2$.
In $3$-dimensional space,the equation $x = k$ represents a plane parallel to the $YZ$-plane.
Therefore,$x = 6$ and $x = 2$ represent two distinct planes,both of which are parallel to the $YZ$-plane.

(D) The formula for the length of the perpendicular from a point $(x_1, y_1, z_1)$ to 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 point $P = (1, 1.5, 2)$ and plane $2x - 2y + 4z + 17 = 0$.
Substituting the values into the formula:
$d = \frac{|2(1) - 2(1.5) + 4(2) + 17|}{\sqrt{2^2 + (-2)^2 + 4^2}}$
$d = \frac{|2 - 3 + 8 + 17|}{\sqrt{4 + 4 + 16}}$
$d = \frac{|24|}{\sqrt{24}}$
$d = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$ units.
Thus,the correct option is $D$.

(B) The equation of the plane is $\frac{x}{2} - \frac{y}{3} - \frac{z}{5} = 1$.
Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,we get the intercepts on the axes as $a = 2, b = -3, c = -5$.
Thus,the coordinates of the points are $A(2, 0, 0)$,$B(0, -3, 0)$,and $C(0, 0, -5)$.
The area of triangle $ABC$ with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),$ and $(x_3, y_3, z_3)$ is given by $\frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2}$.
Substituting the values: $\text{Area} = \frac{1}{2} \sqrt{(2 \times -3)^2 + (-3 \times -5)^2 + (-5 \times 2)^2}$.
$\text{Area} = \frac{1}{2} \sqrt{(-6)^2 + (15)^2 + (-10)^2} = \frac{1}{2} \sqrt{36 + 225 + 100} = \frac{1}{2} \sqrt{361} = \frac{19}{2}$ sq. units.

(B) The equation of the plane is $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$.
The plane cuts the coordinate axes at points $A, B, C$.
Setting $y=0, z=0$,we get $x=2$,so $A = (2, 0, 0)$.
Setting $x=0, z=0$,we get $y=3$,so $B = (0, 3, 0)$.
Setting $x=0, y=0$,we get $z=6$,so $C = (0, 0, 6)$.
The vectors representing the sides are $\vec{AB} = B - A = (-2, 3, 0)$ and $\vec{AC} = C - A = (-2, 0, 6)$.
The area of triangle $ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Calculating the cross product: $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ -2 & 0 & 6 \end{vmatrix} = \hat{i}(18 - 0) - \hat{j}(-12 - 0) + \hat{k}(0 - (-6)) = 18\hat{i} + 12\hat{j} + 6\hat{k}$.
The magnitude is $\sqrt{18^2 + 12^2 + 6^2} = \sqrt{324 + 144 + 36} = \sqrt{504} = \sqrt{36 \times 14} = 6\sqrt{14}$.
Thus,the area is $\frac{1}{2} \times 6\sqrt{14} = 3\sqrt{14}$ sq. units.

(B) Let the point be $P(x_1, y_1, z_1) = (-1, 2, -4)$ and the plane be $ax + by + cz + d = 0$,where $a = 1, b = -1, c = -2, d = 1$.
Let the mirror image be $P'(x', y', z')$.
The formula for the mirror image of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is given by:
$\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = \frac{z' - z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$.
First,calculate the value of $ax_1 + by_1 + cz_1 + d$:
$1(-1) - 1(2) - 2(-4) + 1 = -1 - 2 + 8 + 1 = 6$.
Next,calculate $a^2 + b^2 + c^2$:
$1^2 + (-1)^2 + (-2)^2 = 1 + 1 + 4 = 6$.
Now,substitute these values into the formula:
$\frac{x' - (-1)}{1} = \frac{y' - 2}{-1} = \frac{z' - (-4)}{-2} = -2 \frac{6}{6} = -2$.
Solving for $x', y', z'$:
$x' + 1 = -2 \implies x' = -3$.
$y' - 2 = 2 \implies y' = 4$.
$z' + 4 = 4 \implies z' = 0$.
Thus,the mirror image is $(-3, 4, 0)$.

(A) The foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $P(-1, -1, 2)$.
This point $P$ lies on the plane,and the vector $\vec{OP} = -\hat{i} - \hat{j} + 2\hat{k}$ is normal to the plane.
The equation of a plane passing through a point $\vec{a}$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.
Here,$\vec{a} = -\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n} = -\hat{i} - \hat{j} + 2\hat{k}$.
Substituting these values,we get: $((x+1)\hat{i} + (y+1)\hat{j} + (z-2)\hat{k}) \cdot (-\hat{i} - \hat{j} + 2\hat{k}) = 0$.
$-(x+1) - (y+1) + 2(z-2) = 0$.
$-x - 1 - y - 1 + 2z - 4 = 0$.
$-x - y + 2z - 6 = 0$.
Multiplying by $-1$,we get $x + y - 2z + 6 = 0$.

(A) The normal vectors of the given planes are $\vec{n_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 3\hat{i} + 3\hat{j} + 2\hat{k}$.
Since the required plane is perpendicular to both,its normal vector $\vec{n}$ is given by the cross product $\vec{n_1} \times \vec{n_2}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(2 - 6) + \hat{k}(3 - 6) = -2\hat{i} + 4\hat{j} - 3\hat{k}$.
We can take the normal vector as $\vec{n} = 2\hat{i} - 4\hat{j} + 3\hat{k}$.
The equation of the plane passing through $(x_0, y_0, z_0) = (1, 2, 1)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
$2(x - 1) - 4(y - 2) + 3(z - 1) = 0$.
$2x - 2 - 4y + 8 + 3z - 3 = 0$.
$2x - 4y + 3z + 3 = 0$.

(A) The equation of a plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant:
$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$.
Substituting the points $(2,1,0)$,$(3,-2,4)$,and $(1,-3,3)$:
$\begin{vmatrix} x-2 & y-1 & z-0 \\ 3-2 & -2-1 & 4-0 \\ 1-2 & -3-1 & 3-0 \end{vmatrix} = 0 \implies \begin{vmatrix} x-2 & y-1 & z \\ 1 & -3 & 4 \\ -1 & -4 & 3 \end{vmatrix} = 0$.
Expanding the determinant:
$(x-2)(-9 - (-16)) - (y-1)(3 - (-4)) + z(-4 - 3) = 0$
$(x-2)(7) - (y-1)(7) + z(-7) = 0$
$7x - 14 - 7y + 7 - 7z = 0 \implies 7x - 7y - 7z - 7 = 0 \implies x - y - z - 1 = 0$.
The distance $d$ of a point $(x_0, y_0, z_0)$ from the plane $ax + by + cz + d = 0$ is given by $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.
For the point $(5,3,-1)$ and plane $x - y - z - 1 = 0$:
$d = \frac{|1(5) - 1(3) - 1(-1) - 1|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|5 - 3 + 1 - 1|}{\sqrt{3}} = \frac{2}{\sqrt{3}}$ units.

(A) Let the equation of the plane be $a(x-1) + by + cz = 0$,which simplifies to $ax + by + cz - a = 0$.
Since it passes through $(0,1,0)$,we have $a(0) + b(1) + c(0) - a = 0$,which implies $b = a$.
So the equation is $ax + ay + cz - a = 0$,or $x + y + \frac{c}{a}z - 1 = 0$.
Let $k = \frac{c}{a}$,then the normal vector is $\vec{n_1} = (1, 1, k)$.
The normal to the plane $x+y-3=0$ is $\vec{n_2} = (1, 1, 0)$.
The angle between the planes is $45^{\circ}$,so $\cos(45^{\circ}) = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\frac{1}{\sqrt{2}} = \frac{|1+1+0|}{\sqrt{1^2+1^2+k^2} \sqrt{1^2+1^2+0^2}} = \frac{2}{\sqrt{2+k^2} \sqrt{2}}$.
Squaring both sides: $\frac{1}{2} = \frac{4}{2(2+k^2)}$,which gives $2+k^2 = 4$,so $k^2 = 2$,$k = \pm \sqrt{2}$.
Substituting $k$ back,we get $x + y \pm \sqrt{2}z - 1 = 0$.

(C) The equation of a plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant equation:
$\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}\right| = 0$
Substituting the given points $(3,1,1)$,$(1,2,3)$,and $(-1,4,2)$:
$\left|\begin{array}{ccc} x-3 & y-1 & z-1 \\ 1-3 & 2-1 & 3-1 \\ -1-3 & 4-1 & 2-1 \end{array}\right| = 0$
$\Rightarrow \left|\begin{array}{ccc} x-3 & y-1 & z-1 \\ -2 & 1 & 2 \\ -4 & 3 & 1 \end{array}\right| = 0$
Expanding along the first row:
$(x-3)(1-6) - (y-1)(-2+8) + (z-1)(-6+4) = 0$
$-5(x-3) - 6(y-1) - 2(z-1) = 0$
$-5x + 15 - 6y + 6 - 2z + 2 = 0$
$-5x - 6y - 2z + 23 = 0$
Multiplying by $-1$:
$5x + 6y + 2z - 23 = 0$

(B) Let $M$ be the midpoint of the line segment $PQ$.
The coordinates of $M$ are given by $\left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{5+3}{2}\right) = (2, 3, 4)$.
The direction ratios of the line $PQ$ are $(3-1, 4-2, 3-5) = (2, 2, -2)$.
Since the plane is perpendicular to $PQ$,the normal vector to the plane is $\vec{n} = 2\hat{i} + 2\hat{j} - 2\hat{k}$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the values,we get $2(x-2) + 2(y-3) - 2(z-4) = 0$.
Dividing by $2$,we get $(x-2) + (y-3) - (z-4) = 0$.
$x + y - z - 2 - 3 + 4 = 0$.
$x + y - z - 1 = 0$.

(B) The required plane passes through the point $(1,1,1)$ and is perpendicular to the planes $2x-y-2z=5$ and $3x-6y+2z=7$. The normal vectors of these planes are $\vec{n_1} = 2\hat{i} - \hat{j} - 2\hat{k}$ and $\vec{n_2} = 3\hat{i} - 6\hat{j} + 2\hat{k}$. The normal vector $\vec{n}$ of the required plane is $\vec{n_1} \times \vec{n_2}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 3 & -6 & 2 \end{vmatrix} = \hat{i}(-2-12) - \hat{j}(4+6) + \hat{k}(-12+3) = -14\hat{i} - 10\hat{j} - 9\hat{k}$.
The equation of the plane is $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
$-14(x-1) - 10(y-1) - 9(z-1) = 0$.
$-14x + 14 - 10y + 10 - 9z + 9 = 0$.
$-14x - 10y - 9z + 33 = 0$.
$14x + 10y + 9z = 33$ is incorrect in the provided options,let's recheck the determinant calculation.
$\begin{vmatrix} x-1 & y-1 & z-1 \\ 2 & -1 & -2 \\ 3 & -6 & 2 \end{vmatrix} = (x-1)(-2-12) - (y-1)(4+6) + (z-1)(-12+3) = -14(x-1) - 10(y-1) - 9(z-1) = -14x + 14 - 10y + 10 - 9z + 9 = -14x - 10y - 9z + 33 = 0$.
Thus,$14x + 10y + 9z = 33$.

(D) The equation of a plane passing through $(1, -2, 1)$ is $a(x - 1) + b(y + 2) + c(z - 1) = 0$.
Since the plane is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,its normal vector $\vec{n} = (a, b, c)$ is perpendicular to the normal vectors $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$.
Thus,$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.
We can take the normal vector as $(1, 1, 0)$.
The equation of the plane is $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$,which simplifies to $x + y + 1 = 0$.
The distance of the plane $x + y + 0z + 1 = 0$ from the point $(1, 2, 2)$ is given by $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} = \frac{|1(1) + 1(2) + 0(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|1 + 2 + 1|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ units.

(C) Let the point be $(x_1, y_1, z_1) = (-1, 2, -3)$.
The direction ratios of the two lines are $(a_1, b_1, c_1) = (3, 2, -4)$ and $(a_2, b_2, c_2) = (2, -3, 2)$.
The equation of the plane passing through $(x_1, y_1, z_1)$ and parallel to vectors $\vec{b_1} = (a_1, b_1, c_1)$ and $\vec{b_2} = (a_2, b_2, c_2)$ is given by the determinant:
$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
Substituting the values:
$\begin{vmatrix} x + 1 & y - 2 & z + 3 \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{vmatrix} = 0$
Expanding along the first row:
$(x + 1)(2(2) - (-4)(-3)) - (y - 2)(3(2) - (-4)(2)) + (z + 3)(3(-3) - 2(2)) = 0$
$(x + 1)(4 - 12) - (y - 2)(6 + 8) + (z + 3)(-9 - 4) = 0$
$-8(x + 1) - 14(y - 2) - 13(z + 3) = 0$
$-8x - 8 - 14y + 28 - 13z - 39 = 0$
$-8x - 14y - 13z - 19 = 0$
Multiplying by $-1$,we get $8x + 14y + 13z + 19 = 0$.

(C) The equation of the plane passing through $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$ is given by the determinant equation:
$\begin{vmatrix} x-2 & y-1 & z-0 \\ 4-2 & 1-1 & 1-0 \\ 5-2 & 0-1 & 1-0 \end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix} x-2 & y-1 & z \\ 2 & 0 & 1 \\ 3 & -1 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$(x-2)(0 - (-1)) - (y-1)(2 - 3) + z(-2 - 0) = 0$
$(x-2)(1) - (y-1)(-1) - 2z = 0$
$x - 2 + y - 1 - 2z = 0$
$x + y - 2z = 3$
Let $R'(x, y, z)$ be the image of $R(2, 1, 6)$ with respect to the plane $x + y - 2z - 3 = 0$.
The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is:
$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$
Substituting the values:
$\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-6}{-2} = -2 \frac{2 + 1 - 2(6) - 3}{1^2 + 1^2 + (-2)^2}$
$\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-6}{-2} = -2 \frac{3 - 12 - 3}{1 + 1 + 4} = -2 \frac{-12}{6} = 4$
Equating each part to $4$:
$x - 2 = 4 \Rightarrow x = 6$
$y - 1 = 4 \Rightarrow y = 5$
$z - 6 = -8 \Rightarrow z = -2$
Therefore,the image $R'$ is $(6, 5, -2)$.