Explanation

(a) $m_{1} = 1\ kg,\ m_{2} = 2\ kg,\ m_{3} = 3\ kg$
Position of centre of mass (2, 2, 2,)
$m_{4} = 4\ $kg
New position of centre of mass (0, 0, 0).
For initial position
$X_{CM} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}}{m_{1} + m_{2} + m_{3}}$
$2 = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}}{1 + 2 + 3}$
$m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} = 12$
Similarly, $m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} = 12$
and $m_{1}z_{1} + m_{2}z_{2} + m_{3}z_{3} = 12$
For new position,
$X_{CM}^{'} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$0 = \frac{12 + 4 \times x_{4}}{1 + 2 + 3 + 4}$
$4x_{4} = - 12$
$x_{4} = - 3$
Similarly, $y_{4} = - 3$
$z_{4} = - 3$
$\therefore$ Position of fourth mass (-3, -3, -3)