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 distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=17$ is

Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of the perpendiculars lie on the line:

Let the equation of the plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1$. For $c \in \mathbb{Z}$,if the distance of this plane from the point $(a, -c, c)$ is $\frac{2}{\sqrt{a}}$,then $\frac{a+b}{c}$ is equal to

Let the mirror image of the point $(a, b, c)$ with respect to the plane $3x - 4y + 12z + 19 = 0$ be $(a - 6, \beta, \gamma)$. If $a + b + c = 5$,then $7\beta - 9\gamma$ is equal to

The angle between the line $r = (i + 2j - k) + \lambda (i - j + k)$ and the normal to the plane $r \cdot (2i - j + k) = 4$ is