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 the triangle $$ABC$$, then $$(BC)^2$$ is equal to

Let $$B(\alpha, -2\alpha)$$ and $$C(\beta + 3, \beta)$$

$$\frac{\alpha + \beta + 3 + 1}{3} = 2$$

$$\alpha + \beta + 4 = 6 \implies \alpha + \beta = 2 \implies \beta = 2 - \alpha \quad \text{--- (Equation 1)}$$

$$\frac{-2\alpha - 2 + \beta}{3} = a$$

$$-2\alpha - 2 + \beta = 3a$$

$$-2\alpha - 2 + (2 - \alpha) = 3a$$

$$-3\alpha = 3a \implies \alpha = -a \quad \text{--- (Equation 2)}$$

Since $$B(\alpha, -2\alpha)$$ lies on the line $$x + py = 21a$$:

$$\alpha + p(-2\alpha) = 21a$$

$$\alpha(1 - 2p) = 21a \quad \text{--- (Equation 3)}$$

$$-a(1 - 2p) = 21a$$

$$-1(1 - 2p) = 21 \implies -1 + 2p = 21 \implies 2p = 22 \implies p = 11$$

Since $$C(\beta + 3, \beta)$$ also lies on the line $$x + 11y = 21a$$:

$$(\beta + 3) + 11\beta = 21a$$

$$12\beta + 3 = 21a$$

$$21\alpha + 12\beta + 3 = 0$$

$$21\alpha + 12(2 - \alpha) + 3 = 0$$

$$9\alpha + 27 = 0 \implies \alpha = -3$$

$$\beta = 2 - (-3) = 5$$

$$B(-3, 6) \quad \text{and} \quad C(5 + 3, 5) \implies C(8, 5)$$

$$(BC)^2 = (8 - (-3))^2 + (5 - 6)^2$$

$$(BC)^2 = (11)^2 + (-1)^2 = 121 + 1 = 122$$