The distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ is

The vector equation of the plane containing the lines $r=(\hat{i}+\hat{j})+t(\hat{i}+2 \hat{j}-\hat{k})$ and $r=(\hat{i}+\hat{j})+s(-\hat{i}+\hat{j}-2 \hat{k})$ is

Let $Q$ be the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^{\circ}$,then the area of $\triangle PQR$ is equal to:

The equation of the line passing through $(-4, 1, 3)$,parallel to the plane $x + 2y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z - 2}{-1}$ is

If the line $\frac{2x - 8}{\sin \beta} = \frac{y - \sin \alpha}{1} = \frac{z - 1}{\cos \alpha}$,where $\beta \in R$ and $\sin \beta \neq 1$,lies in the plane $2x - (\sin \beta)y + (\cos \beta)z = k$ for all $\alpha \in R$,then:

If two distinct points $Q$ and $R$ lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$,and $PQ = PR = \sqrt{18}$,where the point $P$ is $(1, -2, 3)$,then the area of the triangle $PQR$ is equal to