Answer
$\tau = 12~N\cdot m$
Work Step by Step
We can write a general expression for the torque about the pivot:
$\tau = r\times F = r~F~sin~\theta$
$r$ is the displacement vector from the pivot to the point where the force is applied
$F$ is the force vector
$\theta$ is the angle between these two vectors
We can find the net torque about the pivot:
$\tau = r_A~F_A~sin~\theta_A-r_B~F_B~sin~\theta_B+r_C~F_C~sin~\theta_C$
$\tau = (8.0~m)(10~N)~sin~135^{\circ}-(4.0~m)~(16~N)~sin~90^{\circ}+(3.0~m)~(19~N)~sin~160^{\circ}$
$\tau = 12~N\cdot m$
