A rod of length 2 units whose one end is(1,0,-1) and other end touches the plane  x- 2y+2z+4=0, then

The rod sweeps out the figure which is a cone. The distance of point A(1,0, -1) from the plane is $\frac{|1-2+4|}{\sqrt{9}}=1\text{ unit }$

The slant height L of the cone is 2 units. Then the radius of the base of the cone is $\sqrt{{\mathrm{l}}^2-1}=\sqrt{4-1}=\sqrt{3}$

Hence, the volume of the cone is $\frac{\pi }{3}(\sqrt{3}{)}^2\cdot 1=\pi \text{ cubic units }$

Area of the circle on the plane which the rod traces is 3$\mathrm{\pi }$. Also, the centre of the circle is Q(x, y, z). 

Then 

$\begin{array}{l}\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-0}{-2}=\frac{\mathrm{z}+1}{2}=\frac{-(1-0-2+4)}{1^2+(-2{)}^2+2^2}\\ \mathrm{or} \mathrm{Q}(\mathrm{x},\mathrm{y},\mathrm{z})\equiv \left(\frac{2}{3},\frac{2}{3},\frac{-5}{3}\right)\end{array}$