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

We have,

Since, $G\left(2,a\right)$, so

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

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

So, $\alpha =-a, \beta =2+a$

So, $B\equiv \left(-a,2a\right), C\equiv \left(5+a,2+a\right)$

Now, $B$ and $C$ lies on the line $x+py=21a$, so

$-a+2pa=21a$

$\Rightarrow p=11$

Also,

$5+a+11\left(2+a\right)=21a$

$\Rightarrow 27=9a\Rightarrow a=3$

So,

$B\equiv \left(-3,6\right), C\equiv \left(8,5\right)$

So,

$B{C}^2=121+1=122$