If the line $r = a + t b$ is parallel to the plane $r = c + l d + m e$, then

The value of $m$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z+m}{2}$ lies in the plane $2x-4y+z=7$ is

Consider the planes $3x - 6y - 2z = 15$ and $2x + y - 2z = 5$.
$STATEMENT-1$ : The parametric equations of the line of intersection of the given planes are $x = 3 + 14t, y = 1 + 2t, z = 15t$ because
$STATEMENT-2$ : The vector $14\hat{i} + 2\hat{j} + 15\hat{k}$ is parallel to the line of intersection of the given planes.

The equation of the plane passing through the point $(2, 5, -3)$ and perpendicular to the planes $x + 2y + 2z = 1$ and $x - 2y + 3z = 4$ is:

The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $\pi_1$ is $(1, 3, 5)$. If $(2, 2, -1), (3, 4, 2), (3, 3, 0)$ are three points on the plane $\pi_2$, then the angle between the planes $\pi_1$ and $\pi_2$ is

The distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ is