Let $P_1$ be the plane $3x - y - 7z = 11$ and $P_2$ be the plane passing through the points $(2, -1, 0)$,$(2, 0, -1)$,and $(5, 1, 1)$. If the foot of the perpendicular drawn from the point $(7, 4, -1)$ on the line of intersection of the planes $P_1$ and $P_2$ is $(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma$ is equal to $............$.

If the equation of the plane passing through the point $(2,-3,4)$ and perpendicular to both the planes $2x-3y+5z=2$ and $x+y+2z=3$ is $x+py+qz=r$,then $r$ is equal to

The equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to the $x$-axis is:

Let $Q$ be the mirror image of the point $P(1, 0, 1)$ with respect to the plane $S: x + y + z = 5$. If a line $L$ passing through $(1, -1, -1)$,parallel to the line $PQ$,meets the plane $S$ at $R$,then $QR^{2}$ is equal to

The equation of the straight line passing through $(1, 2, 3)$ and perpendicular to the plane $x + 2y - 5z + 9 = 0$ is

Let a line $l$ pass through the origin and be perpendicular to the lines $l_1: \overrightarrow{r} = (\hat{i} - 11\hat{j} - 7\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ and $l_2: \overrightarrow{r} = (-\hat{i} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})$. If $P$ is the point of intersection of $l$ and $l_1$,and $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $P$ on $l_2$,then $9(\alpha + \beta + \gamma)$ is equal to: