The center of mass (COM) of a system of particles is calculated as:

$X_{COM} = \frac{\sum_{i} m_i x_i}{\sum_{i} m_i}$

$Y_{COM} = \frac{\sum_{i} m_i y_i}{\sum_{i} m_i} $$$

where $m_i$ is the mass of the $i$-th particle, and $(x_i, y_i)$ are its coordinates.

Let's place the origin at the right angle of the triangle and align the sides along the x and y axes:

• Sphere 1: $(4, 0)$


• Sphere 2: $(0, 4)$


• Sphere 3: $(0, 0)$

Since all spheres have mass $2M$, we can simplify the COM calculations:

$$X_{COM} = \frac{2M \cdot 4 + 2M \cdot 0 + 2M \cdot 0} {2M + 2M + 2M} = \frac{4}{3}$$

$$Y_{COM} = \frac{2M \cdot 0 + 2M \cdot 4 + 2M \cdot 0} {2M + 2M + 2M} = \frac{4}{3}$$

The position vector of the COM is $\left(\frac{4}{3}, \frac{4}{3}\right)$. Its magnitude is:

$$|\vec{r}_{COM}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \frac{4\sqrt{2}}{3}$$

We are given that the magnitude of the position vector is of the form $\frac{4\sqrt{2}}{x}$. Comparing this to our result, we find that $x = 3$.

The value of $x$ is 3.