Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$.

If $\theta$ is the angle between the line $\frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z - 2}{4}$ and the plane $2x + y - 3z + 4 = 0$,then $64 \csc^2 \theta$ is equal to

The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3y-6=0$ are:

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive direction of the $x$-axis,then $\cos \alpha$ is:

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is . . . . . . units.

Let a line with direction ratios $a, -4a, -7$ be perpendicular to the lines with direction ratios $3, -1, 2b$ and $b, a, -2$. If the point of intersection of the line $\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$ and the plane $x - y + z = 0$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to $.......$