The angle between the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and the plane $3x + 2y - 3z = 4$ is ......... $^o$

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 $(\alpha, \beta)$ equals.

The symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

The perpendicular distance from the origin to the plane containing the two lines,$\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$,is

The direction ratios (d.r.s.) of the normal to the plane passing through the origin and the line of intersection of the planes $x+2y+3z=4$ and $4x+3y+2z=1$ are

Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^{2}$ is equal to ..... .