Let $\mathrm{L}_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{\mathrm{z}-6}{1}$ and $\mathrm{L}_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $\mathrm{L}_1$ and $\mathrm{L}_2$ is

The perpendicular distance from the origin to the plane containing the two lines $\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$, is

Let $P(2,1,5)$ be a point in space and $Q$ be a point on the line $\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(-3 \hat{i}+\hat{j}+5 \hat{k})$. Then the value of $\mu$ for which the vector $\overline{\mathrm{PQ}}$ is parallel to the plane $3 x-y+4 z=1$ is

The centroid of tetrahedron with vertices $\mathrm{P}(5,-7,0), \mathrm{Q}(\mathrm{a}, 5,3), \mathrm{R}(4,-6, b)$ and $\mathrm{S}(6, \mathrm{c}, 2)$ is $(4,-3,2)$, then the value of $2 a+3 b+c$ is equal to