Let $P(h,k)$ be a point on the curve $y=x^2+7x+2$, nearest to the line $y=3x-3$. Then the equation of the normal to the curve at $P$ is:

The point $P$ on the curve nearest to the line $y=3x-3$ is where the tangent to the curve is parallel to the given line.Slope of the line $y=3x-3$ is $m=3$.Differentiating the curve:
$$\frac{dy}{dx}=2x+7$$
Setting the slope equal to 3:
$$2h+7=3⟹2h=-4⟹h=-2$$
Finding $k$ by substituting $h$ in the curve equation:
$$k=(-2{)}^2+7(-2)+2=4-14+2=-8$$
So, $P$ is $(-2,-8)$.The slope of the tangent at $P$ is $3$, so the slope of the normal is $m_n=-\frac{1}{3}$.Equation of the normal:
$$y-(-8)=-\frac{1}{3}(x-(-2))$$
$$3(y+8)=-(x+2)$$
$$3y+24=-x-2⟹x+3y+26=0$$