Line and Plane Questions in English

(B) 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 $\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$.
Here,$(x_1, y_1, z_1) = (2, 3, 4)$,$(a_1, b_1, c_1) = (1, 1, -k)$,$(x_2, y_2, z_2) = (1, 4, 5)$,and $(a_2, b_2, c_2) = (k, 2, 1)$.
Substituting these values into the determinant condition:
$\left|\begin{array}{ccc} 1-2 & 4-3 & 5-4 \\ 1 & 1 & -k \\ k & 2 & 1 \end{array}\right| = 0$
$\left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{array}\right| = 0$
Expanding the determinant along the first row:
$-1(1 - (-2k)) - 1(1 - (-k^2)) + 1(2 - k) = 0$
$-1(1 + 2k) - 1(1 + k^2) + 2 - k = 0$
$-1 - 2k - 1 - k^2 + 2 - k = 0$
$-k^2 - 3k = 0$
$k^2 + 3k = 0$
$k(k + 3) = 0$
Therefore,$k = 0$ or $k = -3$.

(B) $1$. Direction Cosines: Since the line makes equal angles with the coordinate axes,its direction ratios are proportional to $(1, 1, 1)$.
$2$. Equation of the Line: The parametric form of the line passing through $P(2, 1, 2)$ with direction ratios $(1, 1, 1)$ is:
$\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-2}{1} = t$
Thus,$x = 2+t, y = 1+t, z = 2+t$.
$3$. Point of Intersection with Plane: Substituting these into the plane equation $2x+y+z=9$:
$2(2+t) + (1+t) + (2+t) = 9$
$4 + 2t + 1 + t + 2 + t = 9$
$4t + 7 = 9 \Rightarrow 4t = 2 \Rightarrow t = \frac{1}{2}$.
$4$. Coordinates of $Q$: Substituting $t = \frac{1}{2}$ back,we get $Q = (2+\frac{1}{2}, 1+\frac{1}{2}, 2+\frac{1}{2}) = (\frac{5}{2}, \frac{3}{2}, \frac{5}{2})$.
$5$. Length of $PQ$: Using the distance formula:
$PQ = \sqrt{(\frac{5}{2}-2)^2 + (\frac{3}{2}-1)^2 + (\frac{5}{2}-2)^2}$
$PQ = \sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ units.

(B) For a line to lie in a plane,every point on the line must satisfy the equation of the plane. The line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z+m}{2}$ passes through the point $(4, 2, -m)$.
Substituting this point into the plane equation $2x - 4y + z = 7$:
$2(4) - 4(2) + (-m) = 7$
$8 - 8 - m = 7$
$-m = 7$
$m = -7$
Additionally,the normal vector of the plane $\vec{n} = (2, -4, 1)$ must be perpendicular to the direction vector of the line $\vec{v} = (1, 1, 2)$.
Checking the dot product: $(2)(1) + (-4)(1) + (1)(2) = 2 - 4 + 2 = 0$.
Since the dot product is $0$,the line is parallel to the plane. Since the point $(4, 2, -m)$ lies on the plane for $m = -7$,the entire line lies in the plane.

(B) The given equation of the line is $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$.
This line passes through the point $P(2, 1, -2)$.
Since the line lies in the plane $x+3y-\alpha z+\beta=0$,the point $P$ must satisfy the plane equation:
$2 + 3(1) - \alpha(-2) + \beta = 0$
$2 + 3 + 2\alpha + \beta = 0$
$2\alpha + \beta = -5$ $(i)$
Also,the direction vector of the line $\vec{v} = (3, -5, 2)$ is perpendicular to the normal vector of the plane $\vec{n} = (1, 3, -\alpha)$.
Therefore,their dot product is zero:
$3(1) + (-5)(3) + 2(-\alpha) = 0$
$3 - 15 - 2\alpha = 0$
$-12 - 2\alpha = 0$
$2\alpha = -12 \implies \alpha = -6$
Substituting $\alpha = -6$ into equation $(i)$:
$2(-6) + \beta = -5$
$-12 + \beta = -5$
$\beta = 7$
Thus,$(\alpha, \beta) = (-6, 7)$.

(D) Since the line lies in the plane,the direction vector of the line is perpendicular to the normal vector of the plane. The direction vector of the line is $\vec{v} = 2\hat{i} + \hat{j} + 3\hat{k}$ and the normal to the plane is $\vec{n} = \ell\hat{i} + m\hat{j} - \hat{k}$.
Thus,$\vec{v} \cdot \vec{n} = 0 \Rightarrow 2\ell + m - 3 = 0$,which gives $2\ell + m = 3$ (Equation $i$).
Also,any point on the line must lie on the plane. The point $(3, -2, -4)$ lies on the line,so it must satisfy the plane equation $\ell x + my - z = 9$.
Substituting the point: $3\ell - 2m - (-4) = 9 \Rightarrow 3\ell - 2m = 5$ (Equation $ii$).
Solving the system of equations:
From $(i)$,$m = 3 - 2\ell$. Substituting into $(ii)$:
$3\ell - 2(3 - 2\ell) = 5 \Rightarrow 3\ell - 6 + 4\ell = 5 \Rightarrow 7\ell = 11 \Rightarrow \ell = \frac{11}{7}$.
Then $m = 3 - 2(\frac{11}{7}) = \frac{21 - 22}{7} = -\frac{1}{7}$.
Finally,$\ell^2 + m^2 = (\frac{11}{7})^2 + (-\frac{1}{7})^2 = \frac{121}{49} + \frac{1}{49} = \frac{122}{49}$.

(A) The position vector of point $Q$ on the line is given by $\vec{q} = (1 - 3\mu)\hat{i} + (-1 + \mu)\hat{j} + (2 + 5\mu)\hat{k}$.
Given point $P$ is $(3, 2, 6)$,so its position vector is $\vec{p} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.
The vector $\vec{PQ} = \vec{q} - \vec{p} = (-3\mu - 2)\hat{i} + (\mu - 3)\hat{j} + (5\mu - 4)\hat{k}$.
The normal vector to the plane $x - 4y + 3z = 1$ is $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$.
Since $\vec{PQ}$ is parallel to the plane,it must be perpendicular to the normal vector $\vec{n}$,so $\vec{PQ} \cdot \vec{n} = 0$.
$(-3\mu - 2)(1) + (\mu - 3)(-4) + (5\mu - 4)(3) = 0$.
$-3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$.
$8\mu - 2 = 0$.
$8\mu = 2 \Rightarrow \mu = \frac{2}{8} = \frac{1}{4}$.

(NONE) Given point $P = (2, 1, 5)$.
Any point $Q$ on the line $\vec{r} = (1, -1, 2) + \mu(-3, 1, 5)$ is given by $Q = (1-3\mu, -1+\mu, 2+5\mu)$.
The vector $\vec{PQ} = Q - P = (1-3\mu-2, -1+\mu-1, 2+5\mu-5) = (-1-3\mu, -2+\mu, -3+5\mu)$.
The normal vector to the plane $3x-y+4z=1$ is $\vec{n} = (3, -1, 4)$.
For $\vec{PQ}$ to be parallel to the plane,$\vec{PQ} \cdot \vec{n} = 0$.
$3(-1-3\mu) - 1(-2+\mu) + 4(-3+5\mu) = 0$.
$-3 - 9\mu + 2 - \mu - 12 + 20\mu = 0$.
$10\mu - 13 = 0$.
$\mu = \frac{13}{10}$.

(A) The equation of the plane containing the two lines is given by the determinant form:
$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
Using the point $(1, 4, -4)$ from the second line and the direction vectors $(3, 5, 7)$ and $(1, 4, 7)$:
$\begin{vmatrix} x-1 & y-4 & z+4 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix} = 0$
Expanding the determinant:
$(x-1)(35-28) - (y-4)(21-7) + (z+4)(12-5) = 0$
$7(x-1) - 14(y-4) + 7(z+4) = 0$
Dividing by $7$:
$(x-1) - 2(y-4) + (z+4) = 0$
$x - 2y + z - 1 + 8 + 4 = 0$
$x - 2y + z + 11 = 0$
The perpendicular distance from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
$d = \frac{|11|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{11}{\sqrt{1+4+1}} = \frac{11}{\sqrt{6}}$ units.

(D) The equation of the family of planes passing through the line of intersection of the planes $x+y+z-1=0$ and $2x+3y+4z-5=0$ is given by $(x+y+z-1) + \lambda(2x+3y+4z-5) = 0$.
Rearranging the terms,we get $(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z - (1+5\lambda) = 0$.
This plane is perpendicular to the plane $x-y+z=0$.
The normal vectors of these two planes are $\vec{n_1} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}$ and $\vec{n_2} = \hat{i} - \hat{j} + \hat{k}$.
Since the planes are perpendicular,their dot product is zero: $\vec{n_1} \cdot \vec{n_2} = 0$.
$(1+2\lambda)(1) + (1+3\lambda)(-1) + (1+4\lambda)(1) = 0$.
$1+2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$.
$1 + 3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into the equation of the plane:
$(1 - \frac{2}{3})x + (1 - 1)y + (1 - \frac{4}{3})z - (1 - \frac{5}{3}) = 0$.
$\frac{1}{3}x + 0y - \frac{1}{3}z + \frac{2}{3} = 0$.
Multiplying by $3$,we get $x - z + 2 = 0$.
The vector equation is $\overline{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$.

(A) Let the given line be $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4} = \lambda$.
Any point on the line is given by $(x, y, z) = (2\lambda+1, -3\lambda+2, 4\lambda-5)$.
Since this point lies on the plane $2x+4y-z=3$,we substitute these coordinates into the plane equation:
$2(2\lambda+1) + 4(-3\lambda+2) - (4\lambda-5) = 3$.
Expanding the terms:
$4\lambda + 2 - 12\lambda + 8 - 4\lambda + 5 = 3$.
Combining the $\lambda$ terms and constants:
$-12\lambda + 15 = 3$.
$-12\lambda = 3 - 15$.
$-12\lambda = -12$.
$\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates:
$x = 2(1) + 1 = 3$.
$y = -3(1) + 2 = -1$.
$z = 4(1) - 5 = -1$.
Thus,the required coordinates are $(3, -1, -1)$.

(A) The line $L$ is the intersection of two planes: $2x + 3y + z = 1$ and $x + 3y + 2z = 2$.
The direction vector $\vec{v}$ of the line $L$ is given by the cross product of the normals to the planes,$\vec{n_1} = (2, 3, 1)$ and $\vec{n_2} = (1, 3, 2)$.
$\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(4-1) + \hat{k}(6-3) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.
We can simplify the direction vector to $\vec{v} = (1, -1, 1)$.
The line $L$ makes an angle $\alpha$ with the positive $X$-axis,which has the direction vector $\vec{u} = (1, 0, 0)$.
The cosine of the angle $\alpha$ is given by $\cos \alpha = \frac{|\vec{v} \cdot \vec{u}|}{|\vec{v}| |\vec{u}|}$.
$\cos \alpha = \frac{|(1)(1) + (-1)(0) + (1)(0)|}{\sqrt{1^2 + (-1)^2 + 1^2} \cdot \sqrt{1^2 + 0^2 + 0^2}} = \frac{1}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$.
Therefore,$\sec \alpha = \frac{1}{\cos \alpha} = \sqrt{3}$.

(B) Since the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,the direction ratios of the line $(3, -5, 2)$ must be perpendicular to the normal of the plane $(1, 3, -\alpha)$.
Therefore,$3(1) + (-5)(3) + 2(-\alpha) = 0$.
$3 - 15 - 2\alpha = 0 \Rightarrow -12 - 2\alpha = 0 \Rightarrow \alpha = -6$.
Also,the point $(2, 1, -2)$ on the line must satisfy the plane equation.
$2 + 3(1) - (-6)(-2) + \beta = 0$.
$2 + 3 - 12 + \beta = 0 \Rightarrow -7 + \beta = 0 \Rightarrow \beta = 7$.
Finally,the value of $\alpha^2 + \alpha\beta + \beta^2 = (-6)^2 + (-6)(7) + (7)^2 = 36 - 42 + 49 = 43$.

(C) The angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} \right|$.
Here,the direction vector of the line is $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$ and the normal vector of the plane is $\vec{n} = \hat{i} - 2\hat{j} - \lambda\hat{k}$.
Given $\theta = \cos^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - \left(\frac{2\sqrt{2}}{3}\right)^2} = \sqrt{1 - \frac{8}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Substituting the values into the formula:
$\frac{1}{3} = \left| \frac{(2)(1) + (1)(-2) + (-2)(-\lambda)}{\sqrt{2^2 + 1^2 + (-2)^2} \sqrt{1^2 + (-2)^2 + (-\lambda)^2}} \right|$
$\frac{1}{3} = \left| \frac{2 - 2 + 2\lambda}{\sqrt{4 + 1 + 4} \sqrt{1 + 4 + \lambda^2}} \right|$
$\frac{1}{3} = \left| \frac{2\lambda}{3 \sqrt{5 + \lambda^2}} \right|$
Squaring both sides:
$\frac{1}{9} = \frac{4\lambda^2}{9(5 + \lambda^2)}$
$5 + \lambda^2 = 4\lambda^2$
$3\lambda^2 = 5$
$\lambda^2 = \frac{5}{3} \Rightarrow \lambda = \sqrt{\frac{5}{3}}$.

(B) Let the line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=k$.
Any point $P$ on the line is given by $(2k+1, 3k-1, 4k+2)$.
Since $P$ lies on the plane $x+2y+3z=15$,we substitute the coordinates of $P$ into the plane equation:
$(2k+1) + 2(3k-1) + 3(4k+2) = 15$.
Expanding the terms: $2k + 1 + 6k - 2 + 12k + 6 = 15$.
Combining like terms: $20k + 5 = 15$,which gives $20k = 10$,so $k = \frac{1}{2}$.
Substituting $k = \frac{1}{2}$ back into the coordinates of $P$:
$P = (2(\frac{1}{2})+1, 3(\frac{1}{2})-1, 4(\frac{1}{2})+2) = (2, \frac{1}{2}, 4)$.
The distance of $P(2, \frac{1}{2}, 4)$ from the origin $(0, 0, 0)$ is $\sqrt{2^2 + (\frac{1}{2})^2 + 4^2} = \sqrt{4 + \frac{1}{4} + 16} = \sqrt{20 + \frac{1}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$ units.

(D) The normal vector $\bar{n_1}$ to plane $P_1$ is given by the cross product of its parallel vectors: $\bar{n_1} = (2 \hat{j}+3 \hat{k}) \times (4 \hat{j}-3 \hat{k}) = -18 \hat{i}$.
The normal vector $\bar{n_2}$ to plane $P_2$ is given by the cross product of its parallel vectors: $\bar{n_2} = (\hat{j}-\hat{k}) \times (3 \hat{i}+3 \hat{j}) = 3 \hat{i}-3 \hat{j}-3 \hat{k}$.
The vector $\bar{A}$ is parallel to the line of intersection,so $\bar{A}$ is perpendicular to both normals: $\bar{A} = \bar{n_1} \times \bar{n_2} = (-18 \hat{i}) \times (3 \hat{i}-3 \hat{j}-3 \hat{k}) = 54 \hat{j}-54 \hat{k}$.
We can simplify $\bar{A}$ to $\hat{j}-\hat{k}$ (or $-\hat{j}+\hat{k}$).
Let $\bar{v} = 2 \hat{i}+\hat{j}-2 \hat{k}$. The angle $\theta$ between $\bar{A}$ and $\bar{v}$ is given by $\cos \theta = \frac{|\bar{A} \cdot \bar{v}|}{|\bar{A}| |\bar{v}|}$.
$\bar{A} \cdot \bar{v} = (0)(2) + (1)(1) + (-1)(-2) = 1 + 2 = 3$.
$|\bar{A}| = \sqrt{0^2+1^2+(-1)^2} = \sqrt{2}$.
$|\bar{v}| = \sqrt{2^2+1^2+(-2)^2} = \sqrt{4+1+4} = 3$.
$\cos \theta = \frac{3}{\sqrt{2} \times 3} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{\pi}{4}$. Since $\frac{\pi}{4}$ is not in the options,we consider the angle between $\bar{A} = -\hat{j}+\hat{k}$ and $\bar{v}$.
$\bar{A} \cdot \bar{v} = (0)(2) + (-1)(1) + (1)(-2) = -1 - 2 = -3$.
$\cos \theta = \frac{|-3|}{\sqrt{2} \times 3} = \frac{1}{\sqrt{2}}$.
Wait,if we take the angle between the vectors themselves (not the acute angle),$\cos \theta = \frac{-3}{3\sqrt{2}} = -\frac{1}{\sqrt{2}}$,which gives $\theta = \frac{3\pi}{4}$.

(A) Let the line be $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = \lambda$.
Any point $P$ on the line is given by $(3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$.
Since this point lies on the plane $x - y + z = 5$,we substitute the coordinates:
$(3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 5$.
$3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 5$.
$11\lambda + 5 = 5$.
$11\lambda = 0 \Rightarrow \lambda = 0$.
Thus,the point of intersection $P$ is $(2, -1, 2)$.
We need to find the distance between $P(2, -1, 2)$ and $Q(-1, -5, -10)$.
$PQ = \sqrt{(-1 - 2)^2 + (-5 - (-1))^2 + (-10 - 2)^2}$.
$PQ = \sqrt{(-3)^2 + (-4)^2 + (-12)^2}$.
$PQ = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ units.

(A) The direction ratios of the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ are $2, 4, 1$.
Since the required line is parallel to this line,its direction ratios are also $2, 4, 1$.
The equation of the line passing through $A(1, 3, 2)$ with direction ratios $2, 4, 1$ is $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{1} = \lambda$.
Any point $B$ on this line can be represented as $(2\lambda + 1, 4\lambda + 3, \lambda + 2)$.
Since point $B$ lies on the plane $3x + y + 2z = 5$,we substitute the coordinates of $B$ into the plane equation:
$3(2\lambda + 1) + (4\lambda + 3) + 2(\lambda + 2) = 5$.
$6\lambda + 3 + 4\lambda + 3 + 2\lambda + 4 = 5$.
$12\lambda + 10 = 5$.
$12\lambda = -5 \Rightarrow \lambda = -\frac{5}{12}$.
Substituting $\lambda = -\frac{5}{12}$ back into the coordinates of $B$:
$x = 2(-\frac{5}{12}) + 1 = -\frac{5}{6} + 1 = \frac{1}{6}$.
$y = 4(-\frac{5}{12}) + 3 = -\frac{5}{3} + 3 = \frac{4}{3}$.
$z = -\frac{5}{12} + 2 = \frac{19}{12}$.
Thus,the coordinates of point $B$ are $(\frac{1}{6}, \frac{4}{3}, \frac{19}{12})$.

(C) Let $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = \lambda$.
Therefore, the coordinates of any point $P$ on the line are given by $P = (3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$.
Since this point $P$ lies on the plane $x - y + z = 16$, we substitute the coordinates into the plane equation:
$(3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 16$.
$3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 16$.
$11\lambda + 5 = 16$.
$11\lambda = 11$, which gives $\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $P$, we get $P = (3(1) + 2, 4(1) - 1, 12(1) + 2) = (5, 3, 14)$.
Now, we find the distance between $P(5, 3, 14)$ and $Q(1, 6, 2)$ using the distance formula:
$PQ = \sqrt{(1-5)^2 + (6-3)^2 + (2-14)^2}$.
$PQ = \sqrt{(-4)^2 + (3)^2 + (-12)^2}$.
$PQ = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \text{ units}$.

(A) The line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is perpendicular to each of the normal vectors $\bar{n}_1 = 3 \hat{i}-\hat{j}+\hat{k}$ and $\bar{n}_2 = \hat{i}+4 \hat{j}-2 \hat{k}$.
Therefore,the line is parallel to the vector $\bar{n}_1 \times \bar{n}_2$.
Calculating the cross product:
$\bar{n}_1 \times \bar{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix}$
$= \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1))$
$= \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1)$
$= -2 \hat{i} + 7 \hat{j} + 13 \hat{k}$.

(B) Let the required ratio be $\lambda:1$. The position vector of the point dividing the line segment joining $\vec{a} = -2\hat{i} + 4\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 8\hat{k}$ in the ratio $\lambda:1$ is given by $\vec{r} = \frac{\lambda\vec{b} + 1\vec{a}}{\lambda+1}$.
Substituting the coordinates,we get $\vec{r} = \frac{\lambda(3\hat{i}-5\hat{j}+8\hat{k}) + (-2\hat{i}+4\hat{j}+7\hat{k})}{\lambda+1} = \frac{(3\lambda-2)\hat{i} + (-5\lambda+4)\hat{j} + (8\lambda+7)\hat{k}}{\lambda+1}$.
Since this point lies on the plane $\vec{r} \cdot (\hat{i}-2\hat{j}+3\hat{k}) = 17$,we substitute the components:
$\frac{(3\lambda-2)(1) + (-5\lambda+4)(-2) + (8\lambda+7)(3)}{\lambda+1} = 17$.
$(3\lambda-2) + (10\lambda-8) + (24\lambda+21) = 17(\lambda+1)$.
$37\lambda + 11 = 17\lambda + 17$.
$20\lambda = 6$.
$\lambda = \frac{6}{20} = \frac{3}{10}$.
Thus,the required ratio is $3:10$.

(D) For the line to lie on the plane,every point on the line must satisfy the equation of the plane.
Since the line passes through the point $(4, 2, k)$,this point must satisfy the plane equation $2x - 4y + z = 7$.
Substituting the coordinates of the point into the plane equation:
$2(4) - 4(2) + k = 7$
$8 - 8 + k = 7$
$k = 7$
Thus,the value of $k$ is $7$.

(A) Given point $P = (3, 2, 6)$.
Point $Q$ on the line is $(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.
The vector $\vec{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2)\hat{j} + (2 + 5\mu - 6)\hat{k} = (-2 - 3\mu)\hat{i} + (-3 + \mu)\hat{j} + (-4 + 5\mu)\hat{k}$.
The normal vector to the plane $x - 4y + 3z = 1$ is $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$.
Since $\vec{PQ}$ is parallel to the plane,$\vec{PQ} \cdot \vec{n} = 0$.
$(-2 - 3\mu)(1) + (-3 + \mu)(-4) + (-4 + 5\mu)(3) = 0$.
$-2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0$.
$8\mu - 2 = 0$.
$8\mu = 2$.
$\mu = \frac{2}{8} = \frac{1}{4}$.

(C) The line is given by $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$. The direction vector of the line is $\vec{b} = (1, 1, 2)$ and the normal to the plane $2x-4y+z=7$ is $\vec{n} = (2, -4, 1)$.
First,check if the line is parallel to the plane by calculating the dot product $\vec{b} \cdot \vec{n} = (1)(2) + (1)(-4) + (2)(1) = 2 - 4 + 2 = 0$. Since the dot product is $0$,the line is parallel to the plane.
For the line to lie in the plane,any point on the line must satisfy the plane equation. Let the point on the line be $(4, 2, k)$.
Substituting this point into the plane equation $2x-4y+z=7$:
$2(4) - 4(2) + k = 7$
$8 - 8 + k = 7$
$k = 7$.

(B) The plane passes through the point $A(0, 7, -7)$ and contains the line passing through $B(-1, 3, -2)$ with direction vector $\vec{v} = -3\hat{i} + 2\hat{j} + \hat{k}$.
The vector $\vec{AB} = (-1-0)\hat{i} + (3-7)\hat{j} + (-2+7)\hat{k} = -\hat{i} - 4\hat{j} + 5\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{AB}$ and $\vec{v}$:
$\vec{n} = \vec{AB} \times \vec{v} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & -4 & 5 \\ -3 & 2 & 1 \end{array}\right| = \hat{i}(-4-10) - \hat{j}(-1+15) + \hat{k}(-2-12) = -14\hat{i} - 14\hat{j} - 14\hat{k}$.
Dividing by $-14$,we get the normal vector $\vec{n}' = \hat{i} + \hat{j} + \hat{k}$.
The equation of the plane passing through $(0, 7, -7)$ is $1(x-0) + 1(y-7) + 1(z+7) = 0$,which simplifies to $x + y + z = 0$.

(C) Given the line $\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}$ lies in the plane $3x-14y+6z+49=0$.
Since the line lies in the plane,every point on the line must satisfy the equation of the plane.
The point $(-1, m, 4)$ lies on the given line.
Substituting this point into the plane equation:
$3(-1) - 14(m) + 6(4) + 49 = 0$
$-3 - 14m + 24 + 49 = 0$
$-14m + 70 = 0$
$14m = 70$
$m = 5$
Thus,the value of $m$ is $5$.

(C) The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$.
Since the line lies in the plane,the normal vector of the plane is perpendicular to the direction vector of the line.
The direction vector of the line is $\vec{v} = (3, -5, 2)$ and the normal vector of the plane is $\vec{n} = (1, 3, -\alpha)$.
Thus,$\vec{v} \cdot \vec{n} = 0$:
$(3)(1) + (-5)(3) + (2)(-\alpha) = 0$
$3 - 15 - 2\alpha = 0$
$-12 - 2\alpha = 0 \Rightarrow \alpha = -6$.
Now,the equation of the plane is $x + 3y + 6z + \beta = 0$.
Since the line lies in the plane,any point on the line must satisfy the plane equation. The point $(2, 1, -2)$ lies on the line.
Substituting $(2, 1, -2)$ into the plane equation:
$2 + 3(1) + 6(-2) + \beta = 0$
$2 + 3 - 12 + \beta = 0$
$-7 + \beta = 0 \Rightarrow \beta = 7$.
Finally,the value of $\alpha \beta = (-6)(7) = -42$.

(A) The angle $\theta$ between a line with direction ratios $(a, b, c)$ and a plane with normal vector $\vec{n} = (a_1, b_1, c_1)$ is given by the formula:
$\sin \theta = \frac{|a a_1 + b b_1 + c c_1|}{\sqrt{a^2 + b^2 + c^2} \sqrt{a_1^2 + b_1^2 + c_1^2}}$
Here,the direction ratios of the line are $(a, b, c) = (2, 1, 2)$ and the normal vector to the plane is $\vec{n} = (6, -2, -3)$.
The magnitude of the direction vector of the line is $\sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
The magnitude of the normal vector to the plane is $\sqrt{6^2 + (-2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
Substituting these values into the formula:
$\sin \theta = \frac{|(2)(6) + (1)(-2) + (2)(-3)|}{3 \times 7} = \frac{|12 - 2 - 6|}{21} = \frac{4}{21}$.
Therefore,$\theta = \sin ^{-1}\left(\frac{4}{21}\right)$.

(A) The angle $\theta$ between a line with direction vector $\bar{b}$ and a plane with normal vector $\bar{n}$ is given by $\sin \theta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}| |\bar{n}|}$.
Given the line $\bar{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$,the direction vector is $\bar{b} = \hat{i}+\hat{j}+\hat{k}$.
Given the plane $\bar{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=5$,the normal vector is $\bar{n} = 2\hat{i}-\hat{j}+\hat{k}$.
Calculate the magnitudes: $|\bar{b}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$ and $|\bar{n}| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{4+1+1} = \sqrt{6}$.
Calculate the dot product: $\bar{b} \cdot \bar{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$.
Substitute into the formula: $\sin \theta = \frac{|2|}{\sqrt{3} \cdot \sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$.
Therefore,$\theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)$.

(A) Let the line be $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = \lambda$.
Any point on the line is given by $(x, y, z) = (\lambda+3, 2\lambda+4, 2\lambda+5)$.
Since this point lies on the plane $x+y+z=2$,we substitute the coordinates into the plane equation:
$(\lambda+3) + (2\lambda+4) + (2\lambda+5) = 2$.
$5\lambda + 12 = 2$.
$5\lambda = -10$,which gives $\lambda = -2$.
Substituting $\lambda = -2$ back into the point coordinates,we get the point of intersection:
$x = -2+3 = 1$,$y = 2(-2)+4 = 0$,$z = 2(-2)+5 = 1$.
The point of intersection is $(1, 0, 1)$.
The distance between $(3, 4, 5)$ and $(1, 0, 1)$ is calculated using the distance formula:
$d = \sqrt{(3-1)^2 + (4-0)^2 + (5-1)^2} = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$ units.

(C) line given by $\bar{r} = \bar{a} + \lambda \bar{b}$ is parallel to a plane $\bar{r} \cdot \bar{n} = d$ if and only if the direction vector of the line $\bar{b}$ is perpendicular to the normal vector of the plane $\bar{n}$.
This implies that the dot product of the vectors $\bar{b}$ and $\bar{n}$ must be zero,i.e.,$\bar{b} \cdot \bar{n} = 0$.
Given $\bar{b} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ and $\bar{n} = 3 \hat{i} - 2 \hat{j} - m \hat{k}$.
Calculating the dot product: $(2 \hat{i} + \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} - 2 \hat{j} - m \hat{k}) = 0$.
$(2)(3) + (1)(-2) + (2)(-m) = 0$.
$6 - 2 - 2m = 0$.
$4 - 2m = 0$.
$2m = 4$.
$m = 2$.

(D) Let $P = (7, 5, 2)$.
The equation of the line passing through $P$ and parallel to the given line $\frac{x-1}{3} = \frac{y-2}{6} = \frac{z+1}{2}$ is given by $\frac{x-7}{3} = \frac{y-5}{6} = \frac{z-2}{2} = r$.
Any point $Q$ on this line can be represented as $(3r+7, 6r+5, 2r+2)$.
Since $Q$ lies on the plane $3x+4y+z-8=0$,we substitute the coordinates of $Q$ into the plane equation:
$3(3r+7) + 4(6r+5) + (2r+2) - 8 = 0$.
Expanding the terms:
$9r + 21 + 24r + 20 + 2r + 2 - 8 = 0$.
Combining like terms:
$35r + 35 = 0 \Rightarrow 35r = -35 \Rightarrow r = -1$.
Substituting $r = -1$ back into the coordinates of $Q$:
$Q = (3(-1)+7, 6(-1)+5, 2(-1)+2) = (4, -1, 0)$.
The distance $PQ$ is calculated using the distance formula:
$PQ = \sqrt{(7-4)^2 + (5-(-1))^2 + (2-0)^2} = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$ units.

(C) The angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|}\right|$.
Here,the direction vector of the line is $\vec{b} = 3\hat{i} + \hat{j}$ and the normal vector to the plane is $\vec{n} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The dot product is $\vec{b} \cdot \vec{n} = (3)(1) + (1)(2) + (0)(3) = 3 + 2 + 0 = 5$.
The magnitudes are $|\vec{b}| = \sqrt{3^2 + 1^2} = \sqrt{10}$ and $|\vec{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.
Substituting these values into the formula: $\sin \theta = \frac{5}{\sqrt{10} \cdot \sqrt{14}} = \frac{5}{\sqrt{140}} = \frac{5}{2\sqrt{35}} = \frac{\sqrt{5} \cdot \sqrt{5}}{2\sqrt{7} \cdot \sqrt{5}} = \frac{\sqrt{5}}{2\sqrt{7}}$.
Thus,$\theta = \sin^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$.

(C) The equation of the line is $\bar{r} = \bar{a} + \lambda \bar{b}$,where $\bar{b} = 2 \hat{\imath} + \hat{\jmath} + 2 \hat{k}$.
The equation of the plane is $\bar{r} \cdot \bar{n} = d$,where $\bar{n} = 3 \hat{\imath} - 2 \hat{\jmath} + m \hat{k}$.
Since the line is parallel to the plane,the direction vector of the line $\bar{b}$ must be perpendicular to the normal vector of the plane $\bar{n}$.
Therefore,$\bar{b} \cdot \bar{n} = 0$.
$(2 \hat{\imath} + \hat{\jmath} + 2 \hat{k}) \cdot (3 \hat{\imath} - 2 \hat{\jmath} + m \hat{k}) = 0$.
$(2)(3) + (1)(-2) + (2)(m) = 0$.
$6 - 2 + 2m = 0$.
$4 + 2m = 0$.
$2m = -4$.
$m = -2$.

(A) Let the given line be $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}=\lambda$.
Any point $P$ on the line is given by $P = (2\lambda+1, -3\lambda+2, 4\lambda-3)$.
Since this point $P$ lies on the plane $2x+4y-z=1$,we substitute the coordinates of $P$ into the plane equation:
$2(2\lambda+1) + 4(-3\lambda+2) - (4\lambda-3) = 1$.
Expanding the terms:
$4\lambda + 2 - 12\lambda + 8 - 4\lambda + 3 = 1$.
Combining the $\lambda$ terms and constants:
$-12\lambda + 13 = 1$.
$-12\lambda = -12$,which gives $\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $P$:
$x = 2(1)+1 = 3$,
$y = -3(1)+2 = -1$,
$z = 4(1)-3 = 1$.
Thus,the point of intersection is $(3, -1, 1)$.

(C) The angle $\theta$ between the line $\bar{r}=\bar{a}+\lambda\bar{b}$ and the plane $\bar{r} \cdot \bar{n}=p$ is given by $\sin \theta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}| |\bar{n}|}$.
Here,$\bar{b} = \hat{i} - \hat{j} + \hat{k}$ and $\bar{n} = 2\hat{i} - \hat{j} + \hat{k}$.
Calculating the dot product: $\bar{b} \cdot \bar{n} = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4$.
Calculating the magnitudes: $|\bar{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|\bar{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
Substituting these values into the formula: $\sin \theta = \frac{4}{\sqrt{3} \cdot \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.
Therefore,$\theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$.

(D) The equation of the line is given by $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$.
Any point on this line can be represented in Cartesian coordinates as $(x, y, z) = (2+\lambda, 1-2\lambda, -4+2\lambda)$.
The $XOY$-plane is defined by the equation $z=0$.
Since the point of intersection lies on the $XOY$-plane,we set the $z$-coordinate to zero:
$-4+2\lambda = 0 \Rightarrow 2\lambda = 4 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ back into the coordinates:
$x = 2+2 = 4$
$y = 1-2(2) = 1-4 = -3$
$z = -4+2(2) = 0$.
Thus,the point of intersection is $(4, -3, 0)$,and its position vector is $4 \hat{i}-3 \hat{j}$.

(A) Let the point on the line be $(x, y, z)$. The equation of the line is $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4} = k$.
From this,we get $x = 2k+1$,$y = -3k+2$,and $z = 4k-3$.
Since this point lies on the plane $2x+4y-z=1$,we substitute these coordinates into the plane equation:
$2(2k+1) + 4(-3k+2) - (4k-3) = 1$
$4k + 2 - 12k + 8 - 4k + 3 = 1$
$-12k + 13 = 1$
$-12k = -12$
$k = 1$.
Substituting $k=1$ back into the coordinate expressions:
$x = 2(1)+1 = 3$
$y = -3(1)+2 = -1$
$z = 4(1)-3 = 1$.
Thus,the required point is $(3, -1, 1)$.

(C) The normal vector to the plane $2x - 3y + 6z - 11 = 0$ is $\vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.
The direction vector of the $X$-axis is $\vec{d} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.
The angle $\theta$ between a line with direction vector $\vec{d}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \frac{|\vec{n} \cdot \vec{d}|}{|\vec{n}| |\vec{d}|}$.
Calculating the dot product: $\vec{n} \cdot \vec{d} = (2)(1) + (-3)(0) + (6)(0) = 2$.
Calculating the magnitudes: $|\vec{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ and $|\vec{d}| = 1$.
Thus,$\sin \theta = \frac{2}{7 \times 1} = \frac{2}{7}$.
Given that $\theta = \sin^{-1}(\alpha)$,we have $\sin^{-1}(\alpha) = \sin^{-1}(\frac{2}{7})$.
Therefore,$\alpha = \frac{2}{7}$.

(B) The given planes are $P_1: x - 2y - z + 5 = 0$ and $P_2: x + y + 3z = 6$.
Their normal vectors are $\vec{N}_1 = \hat{i} - 2\hat{j} - \hat{k}$ and $\vec{N}_2 = \hat{i} + \hat{j} + 3\hat{k}$.
The direction vector $\vec{b}$ of the line of intersection is given by the cross product of the normals:
$\vec{b} = \vec{N}_1 \times \vec{N}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -1 \\ 1 & 1 & 3 \end{vmatrix} = \hat{i}(-6 - (-1)) - \hat{j}(3 - (-1)) + \hat{k}(1 - (-2)) = -5\hat{i} - 4\hat{j} + 3\hat{k}$.
The line passes through $(2, 3, 1)$ and is parallel to $\vec{b} = -5\hat{i} - 4\hat{j} + 3\hat{k}$.
Thus,the equation of the line is $\frac{x-2}{-5} = \frac{y-3}{-4} = \frac{z-1}{3}$.

(D) The direction ratios of the given line are $(3, 4, 5)$.
For a line to be parallel to a plane,the normal vector of the plane must be perpendicular to the direction vector of the line.
If the plane equation is $ax+by+cz=d$,the normal vector is $(a, b, c)$.
The condition for perpendicularity is $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Checking option $D$: $2x+y-2z=0$,the normal vector is $(2, 1, -2)$.
Calculating the dot product: $3(2) + 4(1) + 5(-2) = 6 + 4 - 10 = 0$.
Since the dot product is $0$,the line is parallel to the plane $2x+y-2z=0$.

(A) The given lines are $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K}$ and $\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1}$.
For two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ to be coplanar,the condition is $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
Here,$(x_1, y_1, z_1) = (2, 3, 4)$ and $(x_2, y_2, z_2) = (1, 4, 5)$.
Also,$(a_1, b_1, c_1) = (1, 1, -K)$ and $(a_2, b_2, c_2) = (K, 2, 1)$.
Substituting these values into the determinant:
$\begin{vmatrix} 1-2 & 4-3 & 5-4 \\ 1 & 1 & -K \\ K & 2 & 1 \end{vmatrix} = 0$
$\begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -K \\ K & 2 & 1 \end{vmatrix} = 0$
Expanding the determinant:
$-1(1 - (-2K)) - 1(1 - (-K^2)) + 1(2 - K) = 0$
$-1(1 + 2K) - 1(1 + K^2) + 2 - K = 0$
$-1 - 2K - 1 - K^2 + 2 - K = 0$
$-K^2 - 3K = 0$
$K(K + 3) = 0$
Thus,$K = 0$ or $K = -3$. Since $K = -3$ is not in the options,the correct value is $K = 0$.

(B) The equation of the $XY$-plane is $z = 0$.
Let the ratio in which the $XY$-plane divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ be $k:1$.
The $z$-coordinate of the point of division is given by the section formula: $z = \frac{kz_2 + z_1}{k + 1}$.
Since the point lies on the $XY$-plane,$z = 0$.
Therefore,$0 = \frac{k(-3) + (-5)}{k + 1}$.
This implies $-3k - 5 = 0$,so $3k = -5$,which gives $k = -\frac{5}{3}$.
The negative sign indicates that the division is external.
Thus,the ratio is $5:3$ externally.

(A) The equation of the plane is given by $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$.
Comparing this with the standard form $r \cdot n = d$,we have the normal vector $n = \hat{i}-2 \hat{j}+4 \hat{k}$ and $d = 4$.
The position vector of the point is $a = 2 \hat{i}+\hat{j}-\hat{k}$.
The perpendicular distance $D$ of a point $a$ from the plane $r \cdot n = d$ is given by the formula $D = \frac{|a \cdot n - d|}{|n|}$.
First,calculate the dot product $a \cdot n$:
$a \cdot n = (2 \hat{i}+\hat{j}-\hat{k}) \cdot (\hat{i}-2 \hat{j}+4 \hat{k}) = (2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4$.
Next,calculate the magnitude of the normal vector $|n|$:
$|n| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Now,substitute these values into the distance formula:
$D = \frac{|-4 - 4|}{\sqrt{21}} = \frac{|-8|}{\sqrt{21}} = \frac{8}{\sqrt{21}}$.
Thus,the distance is $\frac{8}{\sqrt{21}}$.

(C) Let the point be $P(1, 3, 4)$ and the plane be $2x - y + z + 3 = 0$. Let $A(x_1, y_1, z_1)$ be the foot of the perpendicular from $P$ to the plane.
The direction ratios of the line $PA$ are $(x_1 - 1, y_1 - 3, z_1 - 4)$.
The normal vector to the plane is $\vec{n} = (2, -1, 1)$.
Since $PA$ is perpendicular to the plane,the direction ratios of $PA$ are proportional to the normal vector:
$\frac{x_1 - 1}{2} = \frac{y_1 - 3}{-1} = \frac{z_1 - 4}{1} = \lambda$
This gives $x_1 = 2\lambda + 1$,$y_1 = -\lambda + 3$,and $z_1 = \lambda + 4$.
Since $A$ lies on the plane $2x - y + z + 3 = 0$,we substitute these coordinates into the plane equation:
$2(2\lambda + 1) - (-\lambda + 3) + (\lambda + 4) + 3 = 0$
$4\lambda + 2 + \lambda - 3 + \lambda + 4 + 3 = 0$
$6\lambda + 6 = 0$
$6\lambda = -6 \implies \lambda = -1$.
Substituting $\lambda = -1$ back into the expressions for $x_1, y_1, z_1$:
$x_1 = 2(-1) + 1 = -1$
$y_1 = -(-1) + 3 = 4$
$z_1 = (-1) + 4 = 3$
Thus,the foot of the perpendicular is $(-1, 4, 3)$.

(D) The given equation of the plane is $5y + 8 = 0$.
The normal vector to the plane is $\vec{n} = (0, 5, 0)$.
The line passing through the origin $(0, 0, 0)$ and perpendicular to the plane has the direction ratios of the normal vector.
Thus,the equation of the line is $\frac{x-0}{0} = \frac{y-0}{5} = \frac{z-0}{0} = \lambda$.
This gives $x = 0$,$y = 5\lambda$,and $z = 0$.
Since the foot of the perpendicular lies on the plane,we substitute these coordinates into the plane equation:
$5(5\lambda) + 8 = 0$
$25\lambda = -8$
$\lambda = -\frac{8}{25}$.
Substituting $\lambda$ back into the coordinates:
$x = 0$,$y = 5(-\frac{8}{25}) = -\frac{8}{5}$,$z = 0$.
Therefore,the coordinates of the foot of the perpendicular are $(0, -\frac{8}{5}, 0)$.

(NONE) The equation of the line is $\frac{x-2}{2}=\frac{y-3}{4}=\frac{z-4}{-2}$.
This line is parallel to the vector $\vec{b} = 2\hat{i} + 4\hat{j} - 2\hat{k}$.
The equation of the plane is $2x - 2y + z = 5$.
The normal vector to the plane is $\vec{n} = 2\hat{i} - 2\hat{j} + \hat{k}$.
The sine of the angle $\theta$ between the line and the plane is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Calculating the dot product: $\vec{b} \cdot \vec{n} = (2)(2) + (4)(-2) + (-2)(1) = 4 - 8 - 2 = -6$.
So,$|\vec{b} \cdot \vec{n}| = |-6| = 6$.
Calculating the magnitudes: $|\vec{b}| = \sqrt{2^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.
$|\vec{n}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,$\sin \theta = \frac{6}{(2\sqrt{6})(3)} = \frac{6}{6\sqrt{6}} = \frac{1}{\sqrt{6}}$.

(A) The plane contains the point $P(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$.
Any point on the line is $Q(3,6,4)$.
The vector along the line is $\vec{v} = \hat{i} + 5\hat{j} + 4\hat{k}$.
The vector joining $P(3,2,0)$ and $Q(3,6,4)$ is $\vec{PQ} = (3-3)\hat{i} + (6-2)\hat{j} + (4-0)\hat{k} = 4\hat{j} + 4\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{v} \times \vec{PQ}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 5 & 4 \\ 0 & 4 & 4 \end{vmatrix} = \hat{i}(20-16) - \hat{j}(4-0) + \hat{k}(4-0) = 4\hat{i} - 4\hat{j} + 4\hat{k}$.
Dividing by $4$,we can take the normal vector as $\vec{n}' = \hat{i} - \hat{j} + \hat{k}$.
The equation of the plane passing through $(3,2,0)$ is $1(x-3) - 1(y-2) + 1(z-0) = 0$.
$x - 3 - y + 2 + z = 0$.
$x - y + z = 1$.