The equation of the plane, passing through the point $(-1,2,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$, is

The co-ordinates of the point where the line through $\mathrm{A}(3,4,1)$ and $\mathrm{B}(5,1,6)$ crosses the $x y$-plane are

The Cartesian equation of a line is $2 x-2=3 y+1=6 z-2$, then the vector equation of the line is

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