d
From the law of consevation of momentum we know that,

\(\begin{array}{l}
{m_1}{u_1} + {m_2}{u_2} + ...... = {m_1}{v_1} + {m_2}{v_2} + ....\\
Given\,{m_1}\, = \,m,{m_2} = 2m\,and\,{m_3} = 3m\\
and\,{u_1} = 3u,\,{u_2} = 2u\,and\,{u_3} = u
\end{array}\)

Let the velocity when the stick \(= \overrightarrow v \)

Then, according to question,

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