Let R be the focus of the parabola y ^ 2 = 20x and the line y = mx...

Given,

$R$ be the focus of the parabola ${y}^{2} = 20x$ and the line $y = mx + c$ intersect the parabola at two points $P and Q$,

And the points $G\left(10, 10\right) be the centroid of the triangle$PQR$,</p><p>Now focus of the parabola${y}^{2} = 20x$will be,$R\left(5, 0\right)$</p><p>And parametric points of$PQ$be \left(5{t}^{2}, 10t\right)$,

Now plotting the diagram we get,

Now finding the centroid of the triangle we get,

For $x -$ coordinate we get,

$⇒ \dfrac{5{{t}_{1}}^{2} + 5{{t}_{2}}^{2} + 5}{3} = 10$

$⇒ {{t}_{1}}^{2} + {{t}_{2}}^{2} = 5 . . . ( i )$

Now for $y -$coordinate we get,

$\dfrac{10\left({t}_{1} + {t}_{2}\right)}{3} = 10$

$⇒ {t}_{1} + {t}_{2} = 3 . . . ( ii )$

Now solving both equations we get, ${t}_{1} = 1, {t}_{2} = 2$

So, points will be, $P ≡ ( 5, 10 ) \text{\text{ and } } Q ≡ \left(20, 20\right)$

Hence, equation of $PQ = y - 10 = \dfrac{10}{15}\left(x - 5\right)</p><p>$⇒ 3y - 30 = 2x - 10$</p><p>$⇒ y = $\dfrac{2}{3}$x + $\dfrac{20}{3}, so on comparing$ with y = mx + c$, we get$c - m = 6$</p><p>Hence,$P{Q}^{2} = 225 + 100 = 325$</p></blockquote>$