Hello Roland2rule.
Let's take a look. The equation of the parabola is $$x=2y^2+5y-17.$$ The equation of the line L is $$L:\ y-y_0=m\left(x-x_0\right)\Leftrightarrow\ \ y-y_0=3\left(x-x_0\right).$$ We just have to find a point that is on the line.
In order to find it we have to solve the quadratic equation $$-5=2y^2+5y-17\ \Leftrightarrow\ \ 2y^2+5y-12=0.$$ The Roots of this equation are given by $$y_{1,2}=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot\left(-12\right)}}{2\cdot2}\ =\ \frac{-5\pm\sqrt{25+96}}{4}=\frac{-5\pm\sqrt{121}}{4}=\ \frac{-5\pm11}{4}.$$ The two solutions are y=6/4=3/2 and y=-16/4=-4.
Since the line intersects the parabola in the upper-left quadrant at x = -5, then we have to select the positive solution.
Therefore, the point of intersection is (-5,3/2). So, the equation of the line is $$L\ :\ y-\frac{3}{2}=3\left(x-\left(-5\right)\right)\ \Leftrightarrow\ \frac{2y-3}{2}=3x+15\ \Leftrightarrow\ 2y-3=6x+30$$ $$\ \Leftrightarrow\ \ -6x+2y-33=0\ \Leftrightarrow\ -3x+y-16.5=0.$$ Hence, the correct option is C.
I hope this explanation may help you.
Feel free to ask me again if you have a doubt.
Regards.
