The equations of the sides AB, BC and CA of a triangle ABC are:
$2\mathrm{x}+\mathrm{y}=0,\mathrm{x}+\mathrm{py}=21\mathrm{a},(\mathrm{a}\ne 0)\text{ and }\mathrm{x}-\mathrm{y}=3$ respectively. Let P(2, a) be the centroid of $\mathrm{△}\mathrm{ABC}$
. Then $(\mathrm{BC}{)}^2$ is equal to

$\begin{array}{l}2\mathrm{x}+\mathrm{y}=0-\cdots --\left(1\right)\\ \mathrm{x}-\mathrm{y}=0-\cdots -\left(2\right)\\ \mathrm{x}+\mathrm{py}=21\mathrm{a}-\cdots \left(3\right)\\ \text{ solving (1) \& (2) }\mathrm{A}(1,-2)\end{array}$

centriod of triangle ABC is 
$\begin{array}{l}\left(\frac{4+\mathrm{s}+\mathrm{t}}{3},\frac{-2-2\mathrm{s}+\mathrm{t}}{3}\right)=(2,\mathrm{a})\\ \Rightarrow \mathrm{s}+\mathrm{t}=2\dots \dots \dots \dots \left(4\right)\\ -2\mathrm{s}+\mathrm{t}=3\mathrm{a}+2\dots \dots \dots \dots (5\end{array}$
Solving (4) 7 95) we get

$\begin{array}{l}\Rightarrow \mathrm{s}=-\mathrm{a},\mathrm{t}=2+\mathrm{a}\\ \mathrm{B}(-\mathrm{a},2\mathrm{a});\mathrm{C}(\mathrm{a}+5,\mathrm{a}+2)\\ \because \mathrm{B},\mathrm{C}\text{ lies on }\mathrm{x}+\mathrm{py}=21\mathrm{a}\\ \Rightarrow -\mathrm{a}+2\mathrm{ap}=21\mathrm{a}\Rightarrow \mathrm{P}=11\\ \text{ and }\mathrm{pa}+2\mathrm{p}=20\mathrm{a}-5\end{array}$
If p = 11
$\begin{array}{l}\therefore 27=9\mathrm{a}\Rightarrow \mathrm{a}=3\\ \therefore \mathrm{B}(-3,6),\mathrm{C}(8,5),(\mathrm{BC}{)}^2=(8+3{)}^2+(5-6{)}^2\\ \therefore \text{ Distance }(\mathrm{BC}{)}^2=122\end{array}$