Question

Let a triangle PQR be such that P and Q lie on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ and are at a distance of 6 units from R(1, 2, 3). If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta PQR$, then $\alpha + \beta + \gamma$ is equal to :

• A. 4
• B. 5
• C. 6
• D. 8

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We find points P and Q on the given line that are 6 units away from R. Then we compute the centroid of triangle PQR and sum its coordinates.
: Key Formula or Approach:
Any point on line: $x = 8\lambda - 3, y = 2\lambda + 4, z = 2\lambda - 1$.
Distance formula: $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.
Centroid $G = (\frac{x_1+x_2+x_3}{3}, \dots)$.
Step 2: Detailed Explanation:
Distance squared from $R(1, 2, 3)$ to point on line is:
$(8\lambda - 4)^2 + (2\lambda + 2)^2 + (2\lambda - 4)^2 = 36$.
$4(4\lambda - 2)^2 + 4(\lambda + 1)^2 + 4(\lambda - 2)^2 = 36$.
$(16\lambda^2 - 16\lambda + 4) + (\lambda^2 + 2\lambda + 1) + (\lambda^2 - 4\lambda + 4) = 9$.
$18\lambda^2 - 18\lambda + 9 = 9 \implies 18\lambda(\lambda - 1) = 0 \implies \lambda = 0, 1$.
For $\lambda = 0$, $P = (-3, 4, -1)$.
For $\lambda = 1$, $Q = (5, 6, 1)$.
Centroid of $\Delta PQR$ with $R(1, 2, 3)$:
$\alpha = \frac{-3 + 5 + 1}{3} = 1$.
$\beta = \frac{4 + 6 + 2}{3} = 4$.
$\gamma = \frac{-1 + 1 + 3}{3} = 1$.
$\alpha + \beta + \gamma = 1 + 4 + 1 = 6$.
Step 3: Final Answer:
The value of $\alpha + \beta + \gamma$ is 6.