Let $x$, $y$, and $z$ be real numbers such that
$$12x - 9y^2 = 7$$
$$6y - 9z^2 = -2$$
$$12z - 9x^2 = 4$$
Find $6x^2 + 9y^2 + 12z^2$.
1. We start with the given equations:
   \[
   12x - 9y^2 = 7
   \]
   \[
   6y - 9z^2 = -2
   \]
   \[
   12z - 9x^2 = 4
   \]

2. Add all three equations:
   \[
   (12x - 9y^2) + (6y - 9z^2) + (12z - 9x^2) = 7 - 2 + 4
   \]
   Simplifying the right-hand side:
   \[
   12x + 6y + 12z - 9x^2 - 9y^2 - 9z^2 = 9
   \]

3. Rearrange the equation:
   \[
   12x + 6y + 12z = 9x^2 + 9y^2 + 9z^2 + 9
   \]
   Divide the entire equation by 3:
   \[
   4x + 2y + 4z = 3(x^2 + y^2 + z^2 + 1)
   \]

4. Rearrange to isolate the quadratic terms:
   \[
   3(x^2 + y^2 + z^2) - 4x - 2y - 4z = -3
   \]

5. Complete the square for each variable:
   \[
   \left(x^2 - \frac{4}{3}x\right) + \left(y^2 - \frac{2}{3}y\right) + \left(z^2 - \frac{4}{3}z\right) = -1
   \]
   \[
   \left(x - \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 + \left(y - \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \left(z - \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 = -1
   \]
   \[
   \left(x - \frac{2}{3}\right)^2 + \left(y - \frac{1}{3}\right)^2 + \left(z - \frac{2}{3}\right)^2 = 0
   \]

6. Since the sum of squares is zero, each square must be zero:
   \[
   x - \frac{2}{3} = 0 \implies x = \frac{2}{3}
   \]
   \[
   y - \frac{1}{3} = 0 \implies y = \frac{1}{3}
   \]
   \[
   z - \frac{2}{3} = 0 \implies z = \frac{2}{3}
   \]

7. Substitute \(x = \frac{2}{3}\), \(y = \frac{1}{3}\), and \(z = \frac{2}{3}\) into the expression \(6x^2 + 9y^2 + 12z^2\):
   \[
   6\left(\frac{2}{3}\right)^2 + 9\left(\frac{1}{3}\right)^2 + 12\left(\frac{2}{3}\right)^2
   \]
   \[
   = 6 \cdot \frac{4}{9} + 9 \cdot \frac{1}{9} + 12 \cdot \frac{4}{9}
   \]
   \[
   = \frac{24}{9} + \frac{9}{9} + \frac{48}{9}
   \]
   \[
   = \frac{24 + 9 + 48}{9}
   \]
   \[
   = \frac{81}{9}
   \]
   \[
   = 9
   \]

The final answer is \(\boxed{9}\)