Let PQR be a triangle with R ( - 1, 4, 2 ) . Suppose M ( 2, 1, 2 )...

Given: 

$\dfrac{x - 2}{0} = \dfrac{y}{2} = \dfrac{z + 3}{- 1}$ and $\dfrac{x - 1}{1} = \dfrac{y + 3}{- 3} = \dfrac{z + 1}{1}$

$⇒ \dfrac{x - 2}{0} = \dfrac{y}{2} = \dfrac{z + 3}{- 1} = k\left(\text{let}\right)$

$⇒ x = 2, y = 2k, z = - k + 3 . . . \left(i\right)$

Also, $\dfrac{x - 1}{1} = \dfrac{y + 3}{- 3} = \dfrac{z + 1}{1} = \lambda\left(\text{let}\right)$

$⇒ x = \lambda + 1, y = - 3\lambda - 3 . z = \lambda - 1 . . . \left(ii\right)$

Using \left(i\right)\text{and}\left(ii\right)$,</p><p>$⇒ 2 = \lambda + 1, 2k = - 3\lambda - 3$</p><p>$⇒ \lambda = 1, 2k = - 3 - 3$</p><p>$⇒ \lambda = 1, k = - 3$</p><p>$⇒ x = 2, y = - 6, z = 0$</p><p>So, the intersection point of two given lines is \left(2, - 6, 0\right)$.

Now, we know that centroid $G$ divides $MR$ (median) in the ratio $1: 2$.

So, by using section formula, $G ≡ \left(\dfrac{1 \times - 1 + 2 \times 2}{3}, \dfrac{1 \times 4 + 2 \times 1}{3}, \dfrac{1 \times 2 + 2 \times 2}{3}\right)$

So, centroid is $G ( 1, 2, 2 )$.

Let $A$ be the point of intersection of given lines.

$⇒ AG = \sqrt{{\left(2 - 1\right)}^{2} + {\left(2 + 6\right)}^{2} + {\left(2 - 0\right)}^{2}}$

$⇒ AG = \sqrt{1 + 64 + 4}$

$⇒ AG = \sqrt{69}$