Let $P:y^2=4a\mathrm{x}, a>0$ be a parabola with focus $S$.Let the tangents to the parabola $P$ make an angle of $\frac{\pi }{4}$ with the line $y=3x+5$ touch the parabola $P$ at $A$ and $B$. Then the value of $a$ for which $A, B$ and $S$ are collinear is:

Lines making angle $\frac{\pi }{4}$ with $y=3x+5$ 

So using formula $\tan \alpha =\frac{m_2-m_1}{1+m_1{m}_2}$, we have slope $-2 \& \frac{1}{2}$.

Which are perpendicular to each-other so, $A, S, B$are collinear for all $a>0$. (by property that perpendicular tangents will lie on directrix and chord of contact will pass through focus, so the points are collinear )