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**Hint:**Centre of mass is a point of a system where whole mass can be considered to be concentrated. We will use the general formula of centre of mass in order to find the value of $\alpha $ . Formula is given as ${X_{C.M}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$ where ${x_1},{x_2}$ are the positions of masses ${m_1},{m_2}$ in a coordinate system.

**Complete step by step answer:**

Since, it’s given that the centre of mass of $1g,2g$ and $3g$ having total mass of the system is $6g$ can be considered to be concentrated at point origin $(0,0,0)$ .

Let us imagine this total mass as ${m_1} = 6g$ with position ${x_1} = {y_1} = {z_1} = 0$ .

Now, imagine mass ${m_2} = 4g$ is placed at position $({x_2},{y_2},{z_2})$ then this combine system has a Centre of mass given as $(1,2,3)$

Now, putting values in formula ${X_{C.M}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$ we get,

$1 = \dfrac{{6(0) + 4({x_2})}}{{10}}$

${x_2} = \dfrac{{10}}{4}$

Similarly, For Y-axis

$2 = \dfrac{{4{y_2}}}{{10}}$

${y_2} = \dfrac{{20}}{4}$

For Z-axis

$3 = \dfrac{{4{z_2}}}{{10}}$

${z_2} = \dfrac{{30}}{4}$

So, given position vector is $\alpha (\hat i + 2\hat j + 3\hat k)$

Putting values of ${x_2} = \dfrac{{10}}{4}$ ${y_2} = \dfrac{{20}}{4}$ ${z_2} = \dfrac{{30}}{4}$ as in vector form we get

Position vector is $\dfrac{5}{2}(\hat i + 2\hat j + 3\hat k)$

On comparing the both position vectors we get, $\alpha = \dfrac{5}{2}$

Hence, the correct option is (B) $\alpha = \dfrac{5}{2}$.

**Note:**Whenever we have given the position of centre of mass of a system having number of individual particles each of having own mass then, the combined total mass of the system can be considered as at a position of their centre of mass and can be taken as a single system having position of centre of mass and mass equals to total mass of the system.

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