The equations of the sides AB, BC and CA of a triangle ABC are: $2x + y = 0$, $x + py = 21a$, $(a \neq 0)$ and $x - y = 3$ respectively. Let $P(2, a)$ be the centroid of $\triangle ABC$. Then $(BC)^2$ is equal to ______.
Step-by-Step Solution
Key Concept: Find vertices A, B, C by intersecting pairs of lines. Use the centroid condition $\left(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}\right) = (2, a)$ to find $\alpha$, $\beta$, $p$, $a$.
$A = (1,-2)$, $B = (-3, 6)$, $C = (8, 5)$. Centroid $= (2, a)$ confirms $a = 3$. $BC = \sqrt{(8-(-3))^2+(5-6)^2} = \sqrt{121+1} = \sqrt{122}$. $(BC)^2 = 122$.
Correct Answer: 122