The point on $\mathrm{y}={\mathrm{x}}^2+7\mathrm{x}+6$ which is closest to the line $\mathrm{y}-3\mathrm{x}+8=0$ is

given $\mathrm{y}={\mathrm{x}}^2+7\mathrm{x}+6$

$\mathrm{p}(\mathrm{x}, \mathrm{y})$ lies on $\mathrm{y}={\mathrm{x}}^2+7\mathrm{x}+6$

$\mathrm{d}=\perp$ distance from p to the line $\mathrm{y}-3\mathrm{x}+8=0$

$$\begin{gathered} =\left|\frac{\mathrm{y}-3\mathrm{x}+8}{\sqrt{1+9}}\right| \\ =\left|\frac{{\mathrm{x}}^2+7\mathrm{x}+6-3\mathrm{x}+8}{\sqrt{10}}\right| \\ {\mathrm{d}}^1=\frac{{\mathrm{x}}^2+4\mathrm{x}+14}{\sqrt{10}} \end{gathered}$$

d is min or max

$$\begin{gathered} \Rightarrow {\mathrm{d}}^1=0 \Rightarrow \mathrm{x}=-2 \\ \mathrm{x}=-2 \\ \mathrm{y}={\mathrm{x}}^2+7\mathrm{x}+6 \\ =4-14+6=-4 \end{gathered}$$

$\therefore$ point $(-2, -4)$