Explanation

(b) The $(x,y,z)$ coordinates of masses $1g,2g,3g$ and $4g$ are
$\left( x_{1} = 0,y_{1} = 0,z_{1} = 0 \right),\left( x_{2} = 0,y_{2} = 0,z_{2} = 0 \right),\ \ $
$\left( x_{3} = 0,y_{3} = 0,z_{3} = 0 \right),$
$\ \ \left( x_{4} = \alpha,y_{4} = 2\alpha,z_{4} = 3\alpha \right)$
$\therefore X_{CM} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$X_{CM} = \frac{1 \times 0 + 2 \times 0 + 3 \times 0 + 4 \times \alpha}{1 + 2 + 3 + 4}$
$1 = \frac{4\alpha}{10}$ or $\alpha = \frac{5}{2}$
The value of $\alpha$ can also be calculated by $Y_{CM}$ and $Z_{CM}$ as shown below
$Y_{CM} = \frac{1 \times 0 + 2 \times 0 + 3 \times 0 + 4 \times 2\alpha}{1 + 2 + 3 + 4}$
$2 = \frac{8\alpha}{10}$ or $\alpha = \frac{20}{8} = \frac{5}{2}$
$Z_{CM} = \frac{1 \times 0 + 2 \times 0 + 3 \times 0 + 4 \times 3\alpha}{1 + 2 + 3 + 4}$
$3 = \frac{12\alpha}{10}$ or $\alpha = \frac{30}{12} = \frac{5}{2}$