If the normals at the points$P,Q,R$on the parabola $y^2=4x$ meet in the point $\left(h,k\right)$. If the centroid and orthocenter of the triangle $PQR$ is $\left(x_1,y_1\right)$ and$\left(x_2,y_2\right)$ then find the value of $3x_1-2x_2$

Let the three feet of the normals be (am₁², -2am₁), (am₂², -2am₂), (am₃², -2am₃)
And since all the three normals pass-through (h, k.) Equation of normal at (am² ,-2am) 
Y = mx - 2am –am³ it passing through (h, k) then am³ + m(2a-h) + k = 0….(A) 
$\begin{array}{l}\Rightarrow {\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3=0\\ {\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_3{\mathrm{m}}_1=\left(\frac{2\mathrm{a}-\mathrm{h}}{\mathrm{a}}\right);{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{m}}_3=\frac{-\mathrm{k}}{\mathrm{a}}\end{array}$
$\therefore$Centroid of triangle PQR is $\left(\frac{\mathrm{a}\sum {\mathrm{m}}_1^2}{3},\frac{-2\mathrm{a}\sum {\mathrm{m}}_1}{3}\right)$
$=\left(\frac{\mathrm{a}\left\{{\left(\sum {\mathrm{m}}_1\right)}^2-2\sum {\mathrm{m}}_1{\mathrm{m}}_2\right\}}{3},0\right)=\left(\frac{2}{3}(\mathrm{h}-2\mathrm{a}),0\right)$
Now equation of he line through P and perpendicular to QR and equation of the line through Q and perpendicular to RP are $\mathrm{y}+2{\mathrm{am}}_1=\left(\frac{{\mathrm{m}}_2+{\mathrm{m}}_3}{2}\right)\left(\mathrm{x}-{\mathrm{am}}_1^2\right)$
$\mathrm{y}+2{\mathrm{am}}_2=\left(\frac{{\mathrm{m}}_3+{\mathrm{m}}_1}{2}\right)\left(\mathrm{x}-{\mathrm{am}}_2^2\right)$
Subtracting (a) & (2) we get 
$\begin{array}{l}2\mathrm{a}\left({\mathrm{m}}_1-{\mathrm{m}}_2\right)=\frac{\mathrm{x}}{2}\left({\mathrm{m}}_2-{\mathrm{m}}_1\right)-\\ \frac{\mathrm{a}}{2}\left\{{\mathrm{m}}_1^2{\mathrm{m}}_2+{\mathrm{m}}_1^2{\mathrm{m}}_3-{\mathrm{m}}_2^2{\mathrm{m}}_3-{\mathrm{m}}_2^2{\mathrm{m}}_1\right\}\\ 2\mathrm{a}\left({\mathrm{m}}_1-{\mathrm{m}}_2\right)=\frac{\mathrm{x}}{2}\left({\mathrm{m}}_2-{\mathrm{m}}_1\right)-\Rightarrow \frac{\mathrm{a}}{2}\left({\mathrm{m}}_1-{\mathrm{m}}_2\right)\sum {\mathrm{m}}_1{\mathrm{m}}_2\\ \Rightarrow 2\mathrm{a}=-\frac{\mathrm{x}}{2}-\frac{\mathrm{a}}{2}\sum {\mathrm{m}}_1{\mathrm{m}}_2\Rightarrow 2\mathrm{a}=-\frac{\mathrm{x}}{2}-\frac{\mathrm{a}}{2}\cdot \frac{(2\mathrm{a}-\mathrm{h})}{\mathrm{a}}\\ \Rightarrow 4\mathrm{a}=-\times -(2\mathrm{a}-\mathrm{h})\Rightarrow \times \mathrm{x}=\mathrm{h}-6\mathrm{a}\end{array}$
Now substituting the values of x in (1) then
$$\begin{gathered} \mathrm{y}+2{\mathrm{am}}_1=\frac{{\mathrm{m}}_2+{\mathrm{m}}_3}{2}+\mathrm{h}-6\mathrm{a}-{\mathrm{am}}_1^2 \\ \begin{array}{l}=-\frac{{\mathrm{m}}_1}{2}\left\{\mathrm{h}-6\mathrm{a}-{\mathrm{am}}_1^2\right\}\\ \mathrm{y}+2{\mathrm{am}}_1=-\frac{{\mathrm{m}}_1\mathrm{h}}{2}+3{\mathrm{am}}_1+\frac{{\mathrm{am}}_1^3}{2}\\ \Rightarrow \mathrm{y}={\mathrm{am}}_1-\frac{{\mathrm{m}}_1\mathrm{h}}{2}+\frac{{\mathrm{am}}_1^3}{2}=\frac{1}{2}\left(2\mathrm{am},-\mathrm{m},\mathrm{h}+{\mathrm{am}}_1^3\right)\\ =\frac{1}{2}(-\mathrm{k})\mathrm{from}\left(\mathrm{A}\right)\end{array} \end{gathered}$$
y = - k/2. Hence orthocenter is (h - 6a, -k/2).