The shortest distance between the line $\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})$ and the plane $\mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5$ is

If the points $A(-1,0,7), B(3,2, t), C(5, k,-2)$ are collinear, then the ratio in which the point $P(t, k-2 t, t+k)$ divides the line segment $B C$ is

The direction cosines $l, m, n$ of two lines are satisfying $3 l+m+5 n=0$ and $6 m n-2 n l+5 l m=0$. If $\theta$ is the angle between those lines then $|\cos \theta|=$

A tetrahedron has vertices $O(0,0,0), A(1,2,1)$, $B(2,1,3), C(-1,1,2)$. If $\theta$ is the angle between the faces $O A B$ and $A B C$, then $\cos \theta=$