Total number of elementary events $=6\times 6\times 6=216$

Clearly ,${\omega }^{{\beta }_1}+{\omega }^{{\beta }_2}+{\omega }^{{\beta }_3}=0$

If one of $r_1,r_2,r_3$ takes value from the set $\left\{3,6\right\}$ ,  other take the value from $\left\{1,4\right\}$ and the third one takes the value from $\left\{2,5\right\}$.

The total number of these ways $=\left({}^2{C}_1 \times {}^{2}C_1 \times {}^2{C}_1\right)\times 3!$

So favourable number of elementary events $=\left({}^2{C}_1 \times {}^{2}C_1 \times {}^2{C}_1\right)\times 3!=48$

Required probability$=\frac{48}{216}=\frac{2}{9}$