The distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$ is

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0$. Then the value of $(\beta-\alpha)$ is equal to

$l, m, n$ are three unit vectors in a right-handed system and $L$ is a line through the points $A, B, C$ whose position vectors are $p l + 7 m - 6 n, 2 l + 5 m - 4 n$ and $l + 4 m - 3 n$ respectively. If the equation of the plane containing $L$ and the point $(-p, p, p+1)$ is $ax + by + cz = 1$, then $p(a+b+c) =$

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x+y+3z=16$ at the point $P$. Let the point $Q$ be the foot of the perpendicular from the point $R(1, -1, -3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$,then $\alpha^2$ is equal to $...........$.

The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2y-z+1=0$ and $3x-y-4z+3=0$ is

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $5y + 8 = 0$.