Three Dimensional Geometry

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 $\mathbf{P}$.