The volume of the cube whose two faces lie on the plane $6x-3y+2z+1=0 $ and $6x-3y+2z+4=0$ is equal to
(a) 27
(b) 27/349
(c) 3/7
(D) 10
My work : the distance between the two plane is$ =\frac{6-3+2+4}{\sqrt{6^2+3^2+2^2}}-\frac{6-3+2+1}{\sqrt{6^2+3^2+2^2}}=\frac{3}{7}$.
So the length of one side is= 3/7 . Hence volume is $=(\frac{3}{7})^3=\frac{27}{343}$ .
Is my work correct ??