We have,

Since, G(2,a), so
31+α+β+3=2 and 3−2−2α+β=a
⇒α+β=2 and −2α+β=3a+2
So, α=−a,β=2+a
So, B≡(−a,2a),C≡(5+a,2+a)
Now, B and C lies on the line x+py=21a, so
−a+2pa=21a
⇒p=11
Also,
5+a+11(2+a)=21a
⇒27=9a⇒a=3
So,
B≡(−3,6),C≡(8,5)
So,
BC2=121+1=122
JEE Main 2023 — Mathematics Coordinate Geometry
The equations of the sides AB,BC&CA of a triangle ABC are 2x+y=0, x+py=21a (a=0) and x−y=3 respectively. Let P(2,a) be the centroid of the triangle ABC, then (BC)2 is equal to
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