The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$. Then the value of $\alpha \beta$ is:

Show that the lines $\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta}$ and $\frac{x-b+c}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}$ are coplanar.

$A$ line with direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x + y + z = 9$ at point $Q$. Find the length of the line segment $PQ$.

$A$ line $l$ passes through the origin and is perpendicular to the lines $l_1 = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$ and $l_2 = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$.
Statement $1$: Line $l$ and $l_2$ are coplanar lines.
Statement $2$: Line $l$ and $l_2$ are intersecting lines.

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:

If the line $x = y = z$ intersects the line defined by the equations $x \sin A + y \sin B + z \sin C - 18 = 0$ and $x \sin 2A + y \sin 2B + z \sin 2C - 9 = 0$,where $A, B, C$ are the angles of a triangle $ABC$,then $80 \left( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \right)$ is equal to $..........$.