The shortest distance between the lines $\overline{r}=(3 \bar{i}-5 \bar{j}+2 \bar{k})+t(4 \bar{i}+3 \bar{j}-\bar{k})$ and $\overline{r}=(\bar{i}+2 \bar{j}-4 \bar{k})+s(6 \bar{i}+3 \bar{j}-2 \bar{k})$ is

Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$. Let $-7 \hat{i}-5 \hat{j}+11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $|\overline{AP}|=12$. Then the position vector of $A$ can be

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

If $(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from a point $P(-1, 2, -1)$ to the line joining the points $A(2, -1, 1)$ and $B(1, 1, -2)$, then $\alpha + \beta + \gamma =$

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$,then the sum of all possible values of $\alpha$ is

One vertex of a rectangular parallelepiped is at the origin $O$ and the lengths of its edges along $x, y$ and $z$ axes are $3, 4$ and $5$ units respectively. Let $P$ be the vertex $(3, 4, 5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to the $z$-axis,not passing through $O$ or $P$,is: