If $P=(2,-3,4)$,$Q=(-1,-4,0)$,and $R=(2,1,0)$ are three points,and $S$ is the foot of the perpendicular drawn from $R$ to the line $PQ$,then the $X$-coordinate of $S$ is:

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 shortest distance between the $z$-axis and the line $x + y + 2z - 3 = 0 = 2x + 3y + 4z - 4$ is

The equation of the plane passing through the intersection of the planes $2x - 5y + z = 3$ and $x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is $x + 3y + 6z = k$,where $k$ is:

$l, m, n$ 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 $p(a+b+c) =$

If the perpendicular distance of the plane passing through the point $Q(1, 0, -1)$ and containing the line $\vec{r} = (\hat{i} - 3\hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ from the origin is $\frac{p}{\sqrt{53}}$, then $p = $