#### Answer

$12.2KN$

#### Work Step by Step

We know that the net torque is supposed to be into page and it is zero
$\implies \sum\tau=m_1g+0.38+m_2g\times 3.98+m_3g\times 8.6-F_2\times 0.76$
This simplifies to:
$F_2=\frac{321\times 9.8\times 0.38+175\times 9.8\times 3.98+64.7\times 8.6}{0.76}$
$F_2=17.7KN$
Now $F_1=17.7KN-(m_1+m_2+m_3)g$
We plug in the known values to obtain:
$F_1=17.7KN-(321+175+64.7)$
$F_1=12.2KN$