Answer
$M_y=154$
$M_x=0$
The center of mass is $(3.3,0)$
Work Step by Step
$$M_y=m_1\cdot x_1+m_{2}\cdot x_2+m_{3}\cdot x_3$$
$$M_y=12\cdot (-3)+15\cdot 2+20\cdot 8=154$$
$$M_x=m_1\cdot y_1+m_{2}\cdot y_2+m_{3}\cdot y_3=0$$
$\overline{x}=\frac{M_y}{m_1+m_2+m_3}=\frac{154}{12+15+20}\approx 3.3$
$\overline{y}=\frac{M_x}{m_1+m_2+m_3}=\frac{0}{12+15+20}=0$
so the center of mass is:
$$(3.3,0)$$
