$A$ plane $P$ contains the line $x+2y+3z+1=0=x-y-z-6$ and is perpendicular to the plane $-2x+y+z+8=0$. Then which of the following points lies on $P$?

Let the plane $P$ pass through the intersection of the planes $2x + 3y - z = 2$ and $x + 2y + 3z = 6$,and be perpendicular to the plane $2x + y - z + 1 = 0$. If $d$ is the distance of $P$ from the point $(-7, 1, 1)$,then $d^2$ is equal to:

At what point does the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersect the plane $2x + y + z = 7$?

The direction ratios (d.r.s.) of the normal to the plane passing through the origin and the line of intersection of the planes $x+2y+3z=4$ and $4x+3y+2z=1$ are

If the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $\ell x+m y-z=9$,then $\ell^2+m^2$ is

Let a plane $P$ contain two lines $\overrightarrow{r} = \hat{i} + \lambda(\hat{i} + \hat{j}), \lambda \in R$ and $\overrightarrow{r} = -\hat{j} + \mu(\hat{j} - \hat{k}), \mu \in R$. If $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from the point $M(1, 0, 1)$ to $P$,then $3(\alpha + \beta + \gamma)$ equals