The shortest distance between the lines $$\frac{x+7}{-6}=\frac{y-6}{7}=z$$ and $$\frac{7-x}{2}=y-2=z-6$$ is :

Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the point $$(2,-5,11)$$, then $$\mathrm{d}^{2}$$ is equal to :

The distance of the point (3, 2, $$-$$1) from the plane $$3x - y + 4z + 1 = 0$$ along the line $${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$$ is equal to :

Let $${{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}}$$ lie on the plane $$px - qy + z = 5$$, for some p, q $$\in$$ R. The shortest distance of the plane from the origin is :