Let ${\mathrm{L}}_1:\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-3}{3},{\mathrm{L}}_2:\frac{\mathrm{x}-1}{-1}=\frac{\mathrm{y}-2}{3}=\frac{3(\mathrm{z}-3)}{5}\text{ and }{\mathrm{L}}_3:\frac{\mathrm{x}-1}{-32}=\frac{\mathrm{y}-2}{-19}=\frac{\mathrm{z}-3}{15}$ be three lines. A plane intersecting these lines at A, B and C respectively such that PA= 2, PB = 3 and PC = 6 where P = (1, 2, 3). If V is the volume of the tetrahedron PABC and d is the perpendicular distance of the plane from the point P then