Answer
$$
\begin{aligned}
& F_R=21.0 \mathrm{kN} \\
& d=3.43 \mathrm{~m}
\end{aligned}
$$
Work Step by Step
Summing the forces along the $y$ axis
$$
\begin{aligned}
+\uparrow\left(F_R\right)_y=\Sigma F_y ; \quad-F_R & =-2(6)-\frac{1}{2}(3)(6) \\
F_R & =21.0 \mathrm{kN} \downarrow
\end{aligned}
$$
Now Summing the moments about point $A$,
$$
\begin{aligned}
\zeta+\left(M_R\right)_A=\Sigma M_A ; \quad-21.0(d) & =-2(6)(3)-\frac{1}{2}(3)(6)(4) \\
d & =3.429 \mathrm{~m}=3.43 \mathrm{~m}
\end{aligned}
$$
