The position vector of a point $P$ is $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{a}=-\hat{\mathbf{i}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are two vectors which determine a plane $\pi$. The equation of a line through $P$ normal to $\mathbf{b}$ and lying on the plane $\pi$ is

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 $|\cos \theta|=$