If planes $x - c y - b z = 0$,$c x - y + a z = 0$ and $b x + a y - z = 0$ pass through a straight line then $a^2 + b^2 + c^2 =$

The foot of the perpendicular drawn from the origin to the plane $x+y+3z-4=0$ is

The equation of the plane passing through the point $(2, 0, 5)$ and parallel to the vectors $\hat{i} - \hat{j} + \hat{k}$ and $3\hat{i} + 2\hat{j} + \hat{k}$ is:

Let $6x - 3y + 2z - 6 = 0$ be the given plane. If $a, b, c$ are the intercepts made by the plane on $X, Y, Z$-axes respectively; $l, m, n$ are the direction cosines of a normal drawn to the plane and $p$ is the perpendicular distance from the origin to the plane, then $|al + bm + cn|=$

The vector equation of the plane which is at a distance of $ \frac{3}{\sqrt{14}} $ from the origin and the normal vector from the origin is $ 2 \hat{i}-3 \hat{j}+\hat{k} $ is:

If the mirror image of the point $(2, 4, 7)$ in the plane $3x - y + 4z = 2$ is $(a, b, c)$,then $2a + b + 2c$ is equal to