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

Let the line $L$ pass through the point $(0,1,2)$,intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$. Then the distance of the point $P(1,-9,2)$ from the line $L$ is

$A$ point $P$ lies on a line passing through $Q(1, -2, 3)$ and is parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$. If $P$ lies on the plane $2x + 3y - 4z + 22 = 0$,then the length of the segment $PQ$ is:

The plane $lx + my = 0$ is rotated by an angle $\alpha$ about its line of intersection with the plane $z = 0$. Find the equation of the plane in its new position.

If the plane $P$ passes through the intersection of two mutually perpendicular planes $2x + ky - 5z = 1$ and $3kx - ky + z = 5$,where $k < 3$,and intercepts a unit length on the positive $x$-axis,then the intercept made by the plane $P$ on the $y$-axis is

The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to a line,whose direction ratios are $2,3,-6$ is :