A plane passes through $(2,1,2)$ and $(1,2,1)$ and parallel to the line $2 x=3 y$ and $\mathrm{z}=1$, then the plane also passes through the point

The coordinates of the foot of the perpendicular drawn from a point $\mathrm{P}(-1,1,2)$ to the plane $2 x-3 y+z-11=0$ are

The lines $\frac{x-3}{1}=\frac{y-2}{1}=\frac{z-5}{-k}$ and $\frac{x-4}{\mathrm{k}}=\frac{y-3}{1}=\frac{\mathrm{z}-3}{2}$ are coplanar, hence $\mathrm{k}=$