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Step 1: Identify Coordinates of the Points
• Point R is given as R(-1, 4, 2).
• M(2, 1, 2) is the midpoint of the segment PQ. Let the coordinates of P be $(x_P, y_P, z_P)$ and Q be $(x_Q, y_Q, z_Q)$.
Step 2: Midpoint Condition for PQ
Because M is the midpoint of PQ:
$ \frac{x_P + x_Q}{2} = 2 \;\; \Longrightarrow \;\; x_P + x_Q = 4, $
$ \frac{y_P + y_Q}{2} = 1 \;\; \Longrightarrow \;\; y_P + y_Q = 2, $
$ \frac{z_P + z_Q}{2} = 2 \;\; \Longrightarrow \;\; z_P + z_Q = 4. $
Step 3: Coordinates of the Centroid G
The centroid G of triangle PQR has coordinates:
$ G = \Bigl(\frac{x_P + x_Q + x_R}{3}, \frac{y_P + y_Q + y_R}{3}, \frac{z_P + z_Q + z_R}{3}\Bigr). $
Using $x_P + x_Q = 4$, $y_P + y_Q = 2$, $z_P + z_Q = 4$, and R(-1, 4, 2):
$ x_G = \frac{4 + (-1)}{3} = \frac{3}{3} = 1, $
$ y_G = \frac{2 + 4}{3} = \frac{6}{3} = 2, $
$ z_G = \frac{4 + 2}{3} = \frac{6}{3} = 2. $
Hence, $G = (1, 2, 2).$
Step 4: Parametric Equations of Both Lines
Line 1: $ \frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{-1} $
From $ \frac{x - 2}{0}, $ we get $ x = 2. $ Let the common ratio for $ \frac{y}{2} $ and $ \frac{z + 3}{-1} $ be $t.$ Then:
$ y = 2t, $
$ z + 3 = -t \;\; \Longrightarrow \;\; z = -3 - t. $
Thus, the parametric form is:
$ (x, y, z) = (2,\, 2t,\, -3 - t). $
Line 2: $ \frac{x - 1}{1} = \frac{y + 3}{-3} = \frac{z + 1}{1} $
Let the common parameter be $s.$ Then:
$ x - 1 = s \;\; \Longrightarrow \;\; x = 1 + s, $
$ y + 3 = -3s \;\; \Longrightarrow \;\; y = -3 - 3s, $
$ z + 1 = s \;\; \Longrightarrow \;\; z = s - 1. $
So, the parametric form is:
$ (x, y, z) = (1 + s,\, -3 - 3s,\, s - 1). $
Step 5: Find the Point of Intersection
At the intersection, the coordinates from both lines must match for some values of $t$ and $s.$
From Line 1 we have $ x = 2.$ Equate it to the $x$ of Line 2:
$ 2 = 1 + s \;\; \Longrightarrow \;\; s = 1. $
Substitute $s = 1$ back into Line 2:
$ x = 1 + 1 = 2, $
$ y = -3 - 3(1) = -6, $
$ z = 1 - 1 = 0. $
Hence, the intersection point is $A = (2, -6, 0).$
Check with Line 1: if $ x = 2,$ then $y = 2t, z = -3 - t.$ Setting $y = -6$ gives $2t = -6 \;\; \Longrightarrow \;\; t = -3,$ and then $z = -3 - (-3) = 0.$ Thus, it checks out.
Step 6: Distance from G to the Intersection Point
Centroid $G = (1, 2, 2)$ and intersection point $A = (2, -6, 0).$ The distance $GA$ is:
$ GA = \sqrt{(1 - 2)^2 + (2 - (-6))^2 + (2 - 0)^2} \\
\quad\,\,= \sqrt{(-1)^2 + (8)^2 + (2)^2} \\
\quad\,\,= \sqrt{1 + 64 + 4} \\
\quad\,\,= \sqrt{69}. $
Final Answer
$ \sqrt{69} $
