The area of the parallelogram with vertices A(1, 2, 3), B(1, 3, a), C(3, 8, 6) and D(3, 7, 3) is $$\sqrt{265}$$ sq. units, then a =

If the lines $\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles, then $\lambda=$

The Cartesian equation of the plane passing through the point A(7, 8, 6) and parallel to the XY plane is

The equation of the plane passing through $$(-2,2,2)$$ and $$(2,-2,-2)$$ and perpendicular to the plane $$9 x-13 y-3 z=0$$ is