The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is

The foot of the perpendicular drawn from the point $(1, 3, 4)$ to the plane $2x - y + z + 3 = 0$ 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:

$A$ line $l$ passes through the origin and is perpendicular to the lines $l_1 = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$ and $l_2 = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$.
Statement $1$: Line $l$ and $l_2$ are coplanar lines.
Statement $2$: Line $l$ and $l_2$ are intersecting lines.

The equation of the plane passing through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

If the lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the equation of the plane containing these lines is: