If we would have drawn a tangent, the slope$(m_1)$ at the point $P$ would have been $m_1 = f'(x) = \frac{-1}{3}$.
As we want to draw the normal, the slope of the normal$(m_2)$ satisfies this property with the tangent : $m_1.m_2 = -1$. In fact, this is true for any two orthogonal. We get $m_2 = 3$.
Now, the equation of the line through this normal will be $y = mx + c$, where $m$ is the slope of this normal which is $3$.
As the line passes through the point $P = (0,5)$. On putting these values in the equation of the line, we get $c = 5$, and thus, the equation of normal through $P$ is $y = 3x + 5$. D is the answer.
