Let $t$ be TNYWR.

The parabola with equation $y=a x^{2}+b x+c$ passes through $(4,0),\left(\frac{t}{3}, 0\right)$, and $(0,60)$. What is the value of $a$ ?
Using that the parabola passes through $(0,60)$ gives $60=a(0)^{2}+b(0)+c$ or $c=60$.

The roots of $a x^{2}+b x+c$ are $x=4$ and $x=\frac{t}{3}$. [We are implicitly assuming that $t \neq 12$ since otherwise the question would not have a unique answer.]

The product of the roots of $a x^{2}+b x+c$ is $\frac{c}{a}=\frac{60}{a}$.

Therefore, $\frac{60}{a}=4 \times \frac{t}{3}$ which can be rearranged to get $a=\frac{3 \times 60}{4 t}=\frac{45}{t}$. Substituting $t=15$ gives $a=3$.