The coordinates of the positions of particles of mass 7, 4 and 10 gm are (1,5, - 3), (2,5,7) and (3, 3, -1)cm respectively. The position of the centre of mass of the system would be

The correct option is (15/7, 85/21, -1/7) cm.

To find the coordinates of the centre of mass of a system of particles, we use the centre of mass formula for a three-dimensional coordinate system.

Given details:
Mass of first particle, $m_1=7\text{g}$ at position $(x_1,y_1,z_1)=(1,5,-3)\text{cm}$
Mass of second particle, $m_2=4\text{g}$ at position $(x_2,y_2,z_2)=(2,5,7)\text{cm}$
Mass of third particle, $m_3=10\text{g}$ at position $(x_3,y_3,z_3)=(3,3,-1)\text{cm}$

First, let's calculate the total mass of the system ($M$):
$M=m_1+m_2+m_3$
$M=7+4+10=21\text{g}$

Now, we compute the coordinates of the centre of mass $(X_{cm},Y_{cm},Z_{cm})$.

1. X-coordinate of the centre of mass ($X_{cm}$):
$X_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{M}$
$X_{cm}=\frac{7\left(1\right)+4\left(2\right)+10\left(3\right)}{21}$
$X_{cm}=\frac{7+8+30}{21}=\frac{45}{21}$
Simplifying the fraction by dividing the numerator and denominator by 3:
$X_{cm}=\frac{15}{7}\text{cm}$

2. Y-coordinate of the centre of mass ($Y_{cm}$):
$Y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{M}$
$Y_{cm}=\frac{7\left(5\right)+4\left(5\right)+10\left(3\right)}{21}$
$Y_{cm}=\frac{35+20+30}{21}=\frac{85}{21}\text{cm}$

3. Z-coordinate of the centre of mass ($Z_{cm}$):
$Z_{cm}=\frac{m_1{z}_1+m_2{z}_2+m_3{z}_3}{M}$
$Z_{cm}=\frac{7(-3)+4\left(7\right)+10(-1)}{21}$
$Z_{cm}=\frac{-21+28-10}{21}=\frac{-3}{21}$
Simplifying the fraction by dividing the numerator and denominator by 3:
$Z_{cm}=-\frac{1}{7}\text{cm}$

Combining the components, we get the position of the centre of mass of the system:
$(X_{cm},Y_{cm},Z_{cm})=(15/7,85/21,-1/7)\text{cm}$