Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle QPR$ is a right angle,then $12a^2$ is equal to

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

Let $P(x_1, y_1, z_1)$ be the foot of the perpendicular drawn from the point $Q(2, -2, 1)$ to the plane $x - 2y + z = 1$. If $d$ is the perpendicular distance from the point $Q$ to the plane and $l = x_1 + y_1 + z_1$,then the value of $l + 3d^2$ is:

The equation of the plane passing through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to the $x$-axis is:

$A$ plane containing the point $(3, 2, 0)$ and the line $\frac{x - 1}{1} = \frac{y - 2}{5} = \frac{z - 3}{4}$ also contains the point

Let $L$ be a line passing through the points $2 \hat{i}+3 \hat{j}+8 \hat{k}$ and $\hat{i}+6 \hat{j}+4 \hat{k}$. Let $P$ be a plane passing through $-5 \hat{i}+19 \hat{j}-14 \hat{k}$ and parallel to the vectors $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-2 \hat{j}+3 \hat{k}$. If $L$ meets the plane $P$ at a point $A$, then the position vector of $A$ is: