The position of center of mass of three masses 2 kg, 3 kg and 15 kg...

Let the midpoint $p$ of the normal bisector be the origin $(0,0)$.

The triangle is isosceles with two sides of $10\text{ m}$ and an included angle of $120^\circ$. The altitude from the $15\text{ kg}$ mass bisects the angle into two $60^\circ$ angles.

The length of the altitude is $10 \cos(60^\circ) = 5\text{ m}$.

Since $p$ is the midpoint of this altitude, the $15\text{ kg}$ mass is at a distance of $2.5\text{ m}$ above $p$, and the base is $2.5\text{ m}$ below $p$.

The distance of the $2\text{ kg} and 3\text{ kg}$ masses from the altitude is $10 \sin(60^\circ) = 5\sqrt{3}\text{ m}$.

The coordinates of the masses with respect to $p$ are:
$m_1 = 15\text{ kg}$ at $(0, 2.5)$
$m_2 = 2\text{ kg}$ at $(-5\sqrt{3}, -2.5)$
$m_3 = 3\text{ kg}$ at $(5\sqrt{3}, -2.5)$

The x-coordinate of the center of mass is:
$x_{cm} = \dfrac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \dfrac{15(0) + 2(-5\sqrt{3}) + 3(5\sqrt{3})}{15 + 2 + 3} = \dfrac{5\sqrt{3}}{20} = \dfrac{\sqrt{3}}{4}$

The y-coordinate of the center of mass is:
$y_{cm} = \dfrac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \dfrac{15(2.5) + 2(-2.5) + 3(-2.5)}{20} = \dfrac{37.5 - 12.5}{20} = \dfrac{25}{20} = 1.25$

The position of the center of mass is \left(\dfrac{\sqrt{3}}{4}, 1.25\right)$.<br><br>Answer: \left(\dfrac{\sqrt{3}}{4}, 1.25\right)$