Calculate the area, in square units, of the region bounded on the right by the line $g(x)=-3 x+27$, on the left by the parabola $f(x)=-(x-4)^{2}+25$, and bounded below by the $x$-axis by integrating with respect to $y$.
Do not include units in your answer.

Step 1 ：First, we need to find the points of intersection of the line and the parabola. We set \(g(x) = f(x)\), which gives us \(-3x + 27 = -(x-4)^2 + 25\).

Step 2 ：Solving this equation, we get \(x^2 - 8x + 16 = 3x - 2\), which simplifies to \(x^2 - 11x + 18 = 0\).

Step 3 ：Factoring this equation, we get \((x - 2)(x - 9) = 0\), so the points of intersection are \(x = 2\) and \(x = 9\).

Step 4 ：Next, we need to find the corresponding y-values for these x-values. Substituting \(x = 2\) into \(g(x)\), we get \(y = -3*2 + 27 = 21\). Substituting \(x = 9\) into \(g(x)\), we get \(y = -3*9 + 27 = 0\).

Step 5 ：So, the points of intersection are \((2, 21)\) and \((9, 0)\).

Step 6 ：Next, we need to set up the integral to find the area. The area is given by the integral from 0 to 21 of \(g^{-1}(y) - f^{-1}(y)\) dy, where \(g^{-1}(y)\) and \(f^{-1}(y)\) are the inverse functions of \(g(x)\) and \(f(x)\) respectively.

Step 7 ：We find that \(g^{-1}(y) = (27 - y) / -3\) and \(f^{-1}(y) = 4 - \sqrt{25 - y}\).

Step 8 ：So, the integral becomes \(\int_{0}^{21} [(27 - y) / -3 - (4 - \sqrt{25 - y})] dy\).

Step 9 ：Evaluating this integral, we get \(-\frac{1}{3} * \frac{1}{2} * y^2 |_{0}^{21} - [4y - \frac{2}{3} * (25 - y)^{3/2}] |_{0}^{21}\).

Step 10 ：Solving this, we get \(-\frac{1}{6} * 21^2 - [4*21 - \frac{2}{3} * (25 - 21)^{3/2}] + \frac{1}{6} * 0^2 + [4*0 - \frac{2}{3} * (25 - 0)^{3/2}]\).

Step 11 ：Solving this, we get \(-49 - 84 + \frac{2}{3} * 8 + \frac{2}{3} * 125\).

Step 12 ：Solving this, we get \(-133 + \frac{2}{3} * 133\).

Step 13 ：Solving this, we get \(-133 + 88.67\).

Step 14 ：Solving this, we get \(-44.33\).

Step 15 ：Since area cannot be negative, we take the absolute value to get the final answer.

Step 16 ：\(\boxed{44.33}\)