Let l be a line which is normal to the curve y = 2x² + x + 2 at a point P on the curve. If the point Q(6, 4) lies on the line l and O is origin, then the area of the triangle OPQ is equal to ______________.

$y=2x^2+x+2\Rightarrow \frac{dy}{dx}=4x+1$
$\text{ Let }\mathrm{P}\text{ be }(\mathrm{h},\mathrm{k})\text{, then normal at }\mathrm{P}\text{ is }y-k=-\frac{1}{4h+1}(x-h)$
$\text{ This passes through }\mathrm{Q}(6,4)$
$$\begin{gathered} \begin{array}{l}\therefore 4-\mathrm{k}=-\frac{1}{4\mathrm{h}+1}(6-\mathrm{h})\\ \Rightarrow (4\mathrm{h}+1)(4-\mathrm{k})+6-\mathrm{h}=0\\ \text{ Also }\mathrm{k}=2{\mathrm{h}}^2+\mathrm{h}+2\\ \therefore (4\mathrm{h}+1)\left(4-2{\mathrm{h}}^2-\mathrm{h}-2\right)+6+\mathrm{h}=0\\ \Rightarrow 4{\mathrm{h}}^3-3{\mathrm{h}}^2+3\mathrm{h}-8=0 \\ \Rightarrow \mathrm{h}=1,\mathrm{k}=5\end{array} \end{gathered}$$
$\text{ Now area of }\mathrm{△}\mathrm{OPQ}\text{ will be }=\frac{1}{2}\left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\end{array}\right|=13$