Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$.

The line passing through the point $2\bar{a}+\bar{b}$ and parallel to the vector $\bar{b}-\bar{c}$ and the plane passing through the point $\bar{a}$ and parallel to the vectors $\bar{b}+\bar{c}$ and $\bar{a}+2\bar{b}-\bar{c}$ intersect at $P$. The position vector of $P$ is

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 $p =$

If the lines $\overrightarrow{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$ and $\overrightarrow{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$ are coplanar,then the distance of the plane containing these two lines from the point $(\alpha, 0, 0)$ is

The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .

Let $L$ be a line obtained from the intersection of two planes $x+2y+z=6$ and $y+2z=4$. If point $P(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $(3,2,1)$ on $L$,then the value of $21(\alpha+\beta+\gamma)$ equals ...... .