From the law of conservation of momentum we know that,
$$
\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2+\ldots .=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2+\ldots .
$$
Given $\mathrm{m}_1=\mathrm{m}, \mathrm{m}_2=2 \mathrm{~m}$ and $\mathrm{m}_3=3 \mathrm{~m}$ and $\mathrm{u}_1=3 \mathrm{u}, \mathrm{u}_2=2 \mathrm{u}$ and $\mathrm{u}_3=\mathrm{u}$
Let the velocity when they stick $=\overrightarrow{\mathrm{v}}$
$$
\text { Then, according to question, }
$$


$$
\begin{aligned}
&m \times 3 u(\hat{i})+2 m \times 2 u\left(-\hat{i} \cos 60^{\circ}-\hat{j} \sin 60^{\circ}\right) \\
&+3 \mathrm{~m} \times \mathrm{u}\left(-\hat{\mathrm{i}} \cos 60^{\circ}+\hat{\mathrm{j}} \sin 60^{\circ}\right) \\
&=(\mathrm{m}+2 \mathrm{~m}+3 \mathrm{~m}) \overrightarrow{\mathrm{v}} \\
&\Rightarrow 3 m u \hat{\mathrm{i}}-4 \mathrm{mu} \frac{\hat{\mathrm{i}}}{2}-4 \mathrm{mu}\left(\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)-3 \mathrm{mu} \frac{\hat{\mathrm{i}}}{2} \\
&+3 m u\left(\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow m u \hat{i}-\frac{3}{2} m u \hat{i}-\frac{\sqrt{3}}{2} m \hat{j}=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow-\frac{1}{2} m u \hat{i}-\frac{\sqrt{3}}{2} m u \hat{j}=6 \mathrm{~m} \mathrm{v} \\
&\Rightarrow \overrightarrow{\mathrm{v}}=\frac{\mathrm{u}}{12}(-\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}}) \\
&
\end{aligned}
$$