1. Given data

• Vertex of triangle: $R(-1,4,2)$
• Midpoint of $PQ$ is $M(2,1,2)$

We need the distance between:

• the centroid of $\triangle PQR$
• the point of intersection of the two given lines.

2. Find the centroid of $\triangle PQR$

If $M$ is the midpoint of $PQ$, then $M=\left(\frac{x_P+x_Q}{2},\frac{y_P+y_Q}{2},\frac{z_P+z_Q}{2}\right).$

So, $x_P+x_Q=2\cdot 2=4,\quad y_P+y_Q=2\cdot 1=2,\quad z_P+z_Q=2\cdot 2=4.$

The centroid $G$ of triangle $PQR$ is $G=\left(\frac{x_P+x_Q+x_R}{3},\frac{y_P+y_Q+y_R}{3},\frac{z_P+z_Q+z_R}{3}\right).$

Substitute $R(-1,4,2)$: $G=\left(\frac{4+(-1)}{3},\frac{2+4}{3},\frac{4+2}{3}\right)=\left(1,2,2\right).$

So the centroid is $G(1,2,2).$

3. Find the intersection point of the two lines

Line 1

Given: $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$

Since denominator of first ratio is $0$, this means $x=2.$

Let the common value be $\lambda$. Then $\frac{y}{2}=\frac{z+3}{-1}=\lambda.$ So, $y=2\lambda,\quad z=-1\lambda-3=-\lambda-3.$

Hence line 1 is $x=2,\quad y=2\lambda,\quad z=-\lambda-3.$

Line 2

Given: $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$

Let common value be $\mu$. Then $x=1+\mu,\quad y=-3-3\mu,\quad z=-1+\mu.$

4. Solve for intersection

At the intersection, coordinates are same.

From $x$: $2=1+\mu \implies \mu=1.$

Then from line 2, $y=-3-3(1)=-6,\quad z=-1+1=0.$

Now check whether this lies on line 1.

For line 1, if $y=2\lambda=-6$, then $\lambda=-3.$ Then $z=-(-3)-3=3-3=0,$ which matches.

So the intersection point is $I(2,-6,0).$

5. Find the distance between $G(1,2,2)$ and $I(2,-6,0)$

Distance formula: $GI=\sqrt{(2-1)^2+(-6-2)^2+(0-2)^2}.$

Compute: