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

Answer: (C)

$\sqrt{69}$

Answer 2

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: $\sqrt{69}$