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

2x + y = 0 ------(1) 
x – y = 0 –----(2)
x + py = 21a  -------(3)
solving (1) & (2) ${\Rightarrow }^{ }$  A (1, -2)
 
centriod of triangle ABC is
$(\frac{4+s+t}{3},\frac{-2-2s+t}{3})=(2,a)$                            $\Rightarrow s+t=\mathrm{2............}\left(4\right)$   
-2s + t = 3a + 2..............(5)
Solving (4) 7 95) we get                         $\Rightarrow s=-a,t=2+a$
 $B(-a,2a);C(a+5,a+2)$
$\because$     B, C lies on x + py = 21a 
$\Rightarrow -a+2ap=21a\Rightarrow P=11$ 
$andpa+2p=20a-5$ 
If  p = 11                                 $\therefore 27=9a\Rightarrow a=3$
$\therefore B(-3,6),C(8,5)$   ,    ${\left(BC\right)}^2={(8+3)}^2+{(5-6)}^2$
 $\therefore$  Distance  ${\left(BC\right)}^2=122$