Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3, a \in R$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals...

If the point of intersection of the lines $r = \hat{i} - 6\hat{j} + (p \sec \alpha) \hat{k} + t(\hat{i} + 2\hat{j} + \hat{k})$ and $r = 4\hat{j} + \hat{k} + \lambda(2\hat{i} + (p \tan \alpha) \hat{j} + 2\hat{k})$ is $8\hat{i} + 8\hat{j} + 9\hat{k}$, (where $0 < \alpha < \frac{\pi}{2}$), then $p =$

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

The image of the point with position vector $\vec{i} + 3\vec{k}$ in the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$ is

The coordinates of the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane $2x - 3y - 6z = 49$ are

The distance of the point $P(1, 2, 1)$ from the plane $2x + y - z = 10$ measured along the line $\frac{x - 5}{1} = \frac{2y - 3}{2} = \frac{z - \frac{5}{2}}{1}$ is