For what value of $k$ do the planes $kx + 4y + z = 0$,$4x + ky + 2z = 0$,and $2x + 2y + z = 0$ intersect in a single line?

$A$ plane ( $\pi$ ) passing through the point $(1, 2, -3)$ is perpendicular to the planes $x + y - z + 4 = 0$ and $2x - y + z + 1 = 0$. If the equation of the plane ( $\pi$ ) is $ax + by + cz + 1 = 0$, then $a^2 + b^2 + c^2 =$

The perpendicular distance from the origin to the plane containing the points $A(1, -2, 1)$, $B(2, -1, -3)$ and $C(0, 1, 5)$ is (in units)

$A$ plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(2, 3, 5)$. Then, the equation of that plane is

The equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$ is:

The point on the plane $2x - 2y + 4z + 5 = 0$ that is nearest to $\left(1, \frac{3}{2}, 2\right)$ is