Let $\mathbf{P}$ be a point on the line $L_1: \frac{x-2}{2}=y+1=\frac{z-1}{2}$ such that its distance from the point $A(2,-1,1)$ is 6 units.

Given that $\boldsymbol{x}$-coordinate of $\mathbf{P}$ is greater than $\mathbf{2}$,

Find the coordinates of point Q on the line $L_2: x-1=\frac{y-2}{2}=\frac{z-2}{2}$ such that $\mathbf{Q}$ is the closest point to $\mathbf{P}$.

The image of a point $P(3,5,3)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is $P^{\prime}(a, b, c)$. Then $a+b+c=$

The equation of a line passing through origin with direction angles $\frac{2 \pi}{3}, \frac{\pi}{4}, \frac{\pi}{3}$ is

Two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect at a point. Then the value of ' $k$ ' is