Let $t$ be TNYWR.

The $y$-coordinate of the vertex of the parabola with equation $y=3 x^{2}+6 \sqrt{m} x+36$ is $t$. What is the value of $m$ ?
We start with the given equation and complete the square:

$y=3 x^{2}+6 \sqrt{m} x+36=3\left(x^{2}+2 \sqrt{m} x+12\right)=3\left((x+\sqrt{m})^{2}-m+12\right)=3(x+\sqrt{m})^{2}+(36-3 m)$

Therefore, the coordinates of the vertex of this parabola are $(-\sqrt{m}, 36-3 m)$.

Since the $y$-coordinate of the vertex is $t$, then $36-3 m=t$ or $3 m=36-t$ and so $m=12-\frac{1}{3} t$.

Since the answer to (b) is 3 , then $t=3$, and so $m=12-\frac{1}{3} t=12-1=11$.

ANSWER: 9,3,11