The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $\pi_1$ is $(1, 3, 5)$. If $(2, 2, -1), (3, 4, 2), (3, 3, 0)$ are three points on the plane $\pi_2$,then the angle between the planes $\pi_1$ and $\pi_2$ is

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1}, b>0$,be $Q(\frac{a}{3}, 0, a+c)$. If $S(\alpha, \beta, \gamma), \alpha > 0$,is the point on $L$ such that the distance of $S$ from the foot of perpendicular $F$ from the point $P$ on $L$ is $2\sqrt{14}$,then $\alpha + \beta + \gamma$ is equal to:

The line $\frac{x - 1}{2} = -(y + 1) = \frac{z}{3}$ and the plane $3x + 2y - z = 5$ intersect at a point. The coordinates of the point are:

If the line passing through the points $(5, 1, a)$ and $(3, b, 1)$ intersects the plane at the point $(0, \frac{17}{2}, -\frac{13}{2})$,then find the values of $a$ and $b$.

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance of $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$.

If the plane passing through the points $\hat{i}+\hat{j}+\hat{k}$,$2\hat{i}-\hat{k}$ and the origin meets the line passing through the points $\hat{i}+3\hat{j}-2\hat{k}$ and $\hat{i}-\hat{j}+3\hat{k}$ at the point $A$,then $A=$