If the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$ meets the plane $x+2y+3z=15$ at the point $P$,then the distance of $P$ from the origin is

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$,$(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are:

Find the length of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ and the coordinates of the foot of the perpendicular.

If the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z + 4}{3}$ lies in the plane $lx + my - z = 9$,then $l^2 + m^2 = \dots$

For $a, b \in \mathbb{Z}$ and $|a - b| \leq 10$,let the angle between the plane $P: ax + y - z = b$ and the line $l: x - 1 = \frac{-y}{1} = z + 1$ be $\cos^{-1}\left(\frac{1}{3}\right)$. If the distance of the point $(6, -6, 4)$ from the plane $P$ is $3\sqrt{6}$,then $a^4 + b^2$ is equal to

$L$ is a line passing through the point $A(1, 0, -3)$ and parallel to a line having direction ratios $0, 1, -2$. $P$ is a point on the line $L$ which is at a minimum distance from the plane $2x + 3y + 5z = 1$. Then,the equation of the plane through $P$ and perpendicular to $AP$ is