By substituting the dimension of given quantities $\left\lbrack ML^{- 1}T^{- 2} \right\rbrack^{x}\left\lbrack MT^{- 3} \right\rbrack^{y}\left\lbrack LT^{- 1} \right\rbrack^{z} = \lbrack MLT\rbrack^{0}$ By comparing the power of $M,L,T$ in both sides $x + y = 0$ ...(i) $- x + z = 0\$ …(ii) $- 2x - 3y - z = 0\$ …(iii) The only values of $x,\ y,\ z$ satisfying (i), (ii) and (iii) corresponds to (b)