The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

$A$ point on the plane that passes through the points $(1, -1, 6)$, $(0, 0, 7)$ and is perpendicular to the plane $x - 2y + z = 6$ is

The length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $\vec r = (\hat i + \hat j) + \lambda (\hat i + 2\hat j - \hat k)$ and $\vec r = (\hat i + \hat j) + \mu (-\hat i + \hat j - 2\hat k)$ is

Let the points on the plane $P$ be equidistant from the points $A(-4, 2, 1)$ and $B(2, -2, 3)$. Then the acute angle between the plane $P$ and the plane $2x + y + 3z = 1$ is

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by:

If the cartesian equation of the plane passing through the point $\hat{i}+2 \hat{j}+\hat{k}$ and parallel to the vectors $2 \hat{i}+3 \hat{j}+\hat{k}$ and $-\hat{i}+2 \hat{j}-3 \hat{k}$ is $a x+b y+c z=1$, then $18(a+b+c)$ is equal to