The given curve is $y=2x^2+x+2$

$\frac{dy}{dx}=4x+1$

Let $P$ be $\left(h,k\right)$, then normal at $P$ is

$y-k=-\frac{1}{4h+1}\left(x-h\right)$

Since this normal passes through $Q\left(6,4\right)$

$\therefore 4-k=-\frac{1}{4h+1}\left(6-h\right)$

$\Rightarrow \left(4h+1\right)\left(4-k\right)+6-h=0$

Also  $k=2h^2+h+2$

$\therefore \left(4h+1\right)\left(4-2h^2-h-2\right)+6+h=0$

$\Rightarrow 4h^3-3h^2+3h-8=0$

$\Rightarrow h=1,k=5$

So $P\equiv \left(1,5\right)$

Now area of $∆OPQ$ will be $=\frac{1}{2}\left|\begin{array}{lll}1& 0& 0\\ 1& 1& 5\\ 1& 6& 4\end{array}\right|=13$