Let PQR be a triangle with R(-1, 4, 2). Suppo | vedclass

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(C) Let the vertices of the triangle be $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. The midpoint $M$ of $PQ$ is given by $(\frac{x_1+x_2}{2}, \frac{y_1

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$\sqrt{69}$

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(C) Let the vertices of the triangle be $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. The midpoint $M$ of $PQ$ is given by $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}) = (2, 1, 2)$.The centroid $G$ of $	riangle PQR$ is $(\frac{x_1+x_2-1}{3}, \frac{y_1+y_2+4}{3}, \frac{z_1+z_2+2}{3})$.Substituting the values from $M$,we get $G = (\frac{4-1}{3}, \frac{2+4}{3}, \frac{4+2}{3}) = (1, 2, 2)$.To find the intersection point $A$ of the lines,let $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1} = k_1$. Then $x = 2, y = 2k_1, z = -k_1-3$.Let $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1} = k_2$. Then $x = k_2+1, y = -3k_2-3, z = k_2-1$.Equating $x$: $2 = k_2+1 \implies k_2 = 1$.Then $y = -3(1)-3 = -6$ and $z = 1-1 = 0$.Checking in the first line: $y = 2k_1 = -6 \implies k_1 = -3$. Then $z = -(-3)-3 = 0$. The point $A$ is $(2, -6, 0)$.The distance $AG = \sqrt{(2-1)^2 + (-6-2)^2 + (0-2)^2} = \sqrt{1^2 + (-8)^2 + (-2)^2} = \sqrt{1 + 64 + 4} = \sqrt{69}$.

Let $PQR$ be a triangle with $R(-1, 4, 2)$. Suppose $M(2, 1, 2)$ is the midpoint of $PQ$. The distance of the centroid of $\triangle PQR$ from the point of intersection of the lines $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$ and $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$ is

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