Question

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

• A. 69
• B. 9
• C. \(\sqrt{69}\)
• D. \(\sqrt{99}\)

Answer 1

The Correct Option is C

Step 1: Find the Centroid \(G\) of \(\triangle PQR\)

Since \(M(2, 1, 2)\) is the midpoint of \(PQ\) and \(R(-1, 4, 2)\) is the third vertex, the centroid \(G\) divides \(MR\) in the ratio \(1 : 2\). Using the section formula to find \(G\):

\[ G = \left(\frac{1 \cdot (-1) + 2 \cdot 2}{1 + 2}, \frac{1 \cdot 4 + 2 \cdot 1}{1 + 2}, \frac{1 \cdot 2 + 2 \cdot 2}{1 + 2}\right) = (1, 2, 2) \]

Step 2: Find the Point of Intersection \(A\) of the Given Lines

Solving the parametric equations of the lines, we find the point of intersection \(A\) to be:

\(A = (2, -6, 0)\)

Step 3: Calculate the Distance \(AG\)

Using the distance formula between points \(G(1, 2, 2)\) and \(A(2, -6, 0)\):

\[ AG = \sqrt{(2 - 1)^2 + (-6 - 2)^2 + (0 - 2)^2} = \sqrt{1 + 64 + 4} = \sqrt{69} \]

So, the correct answer is: \(\sqrt{69}\)