Since the body is in equillibrium so we conclude Vector $F_{\text {net }}=0$ and torque about any point is zero i.e., $\vec{\tau}_{\text {net }}=0$



Let us assume that we apply $F$ force downward at $A$ angle $\theta$ from the horizontal, at $x$ distance from $B$
$\begin{array}{ll}
\text { From } & \vec{F}_{\text {net }}=0 \\
\Rightarrow & F_{\text {net }} x=0 \\
& F_2=8 \mathrm{~N} \\
\text { From } & F_{\text {net } y}=0 \Rightarrow 5+6=F_1+3 \\
\Rightarrow & F_1=8 \mathrm{~N}
\end{array}$

If body is in equillibrium then torque about point $B$ is zero.
$\begin{array}{ll}
\Rightarrow \quad & 3 \times 5+F_1 \cdot x-5 \times 10=0 \\
\Rightarrow \quad & 15+8 x-50=0 \\
& x=\frac{35}{9} \Rightarrow \quad x=4.375 \mathrm{~cm}
\end{array}$