Answer:
[tex](-1.7,\, -3.8,\, 1.5)[/tex].
Explanation:
If that there are [tex]n[/tex] point mass in space, the position vector of the center of mass would be the average of the position vectors [tex]\vec{x}_{1},\, \dots,\, \vec{x}_{n}[/tex] of these objects, weighted according to the mass [tex]m_{1},\, \dots,\, m_{n}[/tex] of each object:
[tex]\displaystyle \frac{m_{1}\, \vec{x}_{1} + \cdots + m_{n}\, \vec{x}_{n}}{m_{1} + \cdots + m_{n}}[/tex].
In this question:
The position vector of the center of mass of these objects would be:
[tex]\displaystyle \frac{m_{1}\, \vec{x}_{1} + m_{2}\, \vec{x}_{2} + m_{3}\, \vec{x}_{3}}{m_{1} + m_{2} + m_{3}}[/tex].
[tex]\begin{aligned}& \frac{(2.0\; {\rm kg})\, \begin{bmatrix}-5.0 \\ -10 \\ -2.0\end{bmatrix} + (5.0\; {\rm kg})\, \begin{bmatrix}1.0 \\ -3.0 \\ 5.0\end{bmatrix} + (3.0\; {\rm kg})\, \begin{bmatrix}-4.0 \\ -1.0 \\ -2.0\end{bmatrix}}{2.0\; {\rm kg} + 5.0\; {\rm kg} + 3.0\; {\rm kg}}} \\ =\; & \frac{1}{10}\, \begin{bmatrix}-17 \\ -38 \\ 15\end{bmatrix} \\ =\; & \begin{bmatrix}-1.7 \\ -3.8 \\ 1.5\end{bmatrix}\end{aligned}[/tex].
In other words, the center of mass of these three objects would be at [tex](-1.7,\, -3.8,\, 1.5)[/tex].
