First, replace the couple with an equivalent couple,
\(M_3\text{,}\) the magnitude of which is
\begin{align*}
M_3 \amp = F d_\perp\\
\amp = F d \sin \ang{60}\\
\amp = \Nm{69.3}
\end{align*}
By observation, this is a counter-clockwise moment, as is
\(M_2\text{.}\) \(M_1\) is clockwise. Summing the scalar magnitudes gives the resultant moment. The signs of the terms are assigned according to the sign convention: positive if counter-clockwise, negative if clockwise.
\begin{align*}
M_R \amp = \Sigma M\\
\amp = M_1 + M_2 + M_3\\
\amp = - \Nm{400} + \Nm{200} + \Nm{69.3}\\
\amp = -\Nm{130.7}\\
\vec{M_R} \amp = \Nm{130.7} \text{ clockwise}
\end{align*}