The point on the parabola y = x² + 7x + 2 which is closest to the line y = 3x – 3 is

y = x²+ 7x + 2
Let point be$(\mathrm{\alpha },\mathrm{\beta })\Rightarrow \mathrm{\beta }={\mathrm{\alpha }}^2+7\mathrm{\alpha }+2$
$S=\frac{|3\alpha -\beta -3|}{\sqrt{9+1}}=\frac{\left|3\alpha -\beta -3\right|}{\sqrt{10}}$
  $$\begin{gathered} \mathrm{let} \mathrm{f}\left(\mathrm{\alpha }\right)=\frac{3\mathrm{\alpha }-\mathrm{\beta }-3}{\sqrt{10}} \\ \Rightarrow f\left(\alpha \right) =\frac{-{\alpha }^2-4\alpha -5}{\sqrt{10}} \\ \Rightarrow {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{\alpha }\right)=0 \\ \Rightarrow \alpha =-2 \\ \Rightarrow \beta =-8 \end{gathered}$$