Answer
$d = 1.3~m$
Work Step by Step
To find $d$, we can consider the torque about a rotation axis located at the origin.
There is a clockwise torque from $F_2$ and a clockwise torque from $F_3$. There is an opposing counterclockwise torque from $F_v$.
In part (b), we found that $F_v = 30~N$
We can find $d$:
$\tau_{net} = 0$
$d~F_v-b~F_2-a~F_3 = 0$
$d~F_v = b~F_2+a~F_3$
$d = \frac{b~F_2+a~F_3}{F_v}$
$d = \frac{(3.0~m)(10~N)+(2.0~m)(5.0~N)}{30~N}$
$d = 1.3~m$
