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

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:

The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then the value of $\alpha^2+\alpha\beta+\beta^2$ is

If a plane $x+y+z-5=0$ intersects the line joining $A(1,1,1)$ and $B(2,2,2)$ at $P$, then $AP: PB=$

$l, m, n$ are three unit vectors in a right-handed system and $L$ is a line through the points $A, B, C$ whose position vectors are $p l + 7 m - 6 n, 2 l + 5 m - 4 n$ and $l + 4 m - 3 n$ respectively. If the equation of the plane containing $L$ and the point $(-p, p, p+1)$ is $ax + by + cz = 1$, then $p(a+b+c) =$