Let the area of the region enclosed by the curves $$y = 3x$$, $$2y = 27 - 3x$$ and $$y = 3x - x\sqrt{x}$$ be $$A$$. Then $$10A$$ is equal to

Find intersection points.

1. $$y = 3x$$ and $$2y = 27 - 3x$$:

$$2(3x) = 27 - 3x \implies 9x = 27 \implies x = 3, y = 9$$.

2. $$y = 3x$$ and $$y = 3x - x\sqrt{x}$$:

$$3x = 3x - x^{3/2} \implies x^{3/2} = 0 \implies x = 0, y = 0$$.

3. $$2y = 27 - 3x$$ and $$y = 3x - x\sqrt{x}$$:

By testing values, at $$x=9$$: $$y = 3(9) - 9(3) = 0$$. And $$2y = 27 - 3(9) = 0$$. So $$x=9, y=0$$.

The area $$A$$ is bounded above by the two lines and below by the curve.

$$A = \int_0^3 (3x - (3x - x^{3/2})) dx + \int_3^9 (\frac{27-3x}{2} - (3x - x^{3/2})) dx$$

Simplifying:

$$A = \int_0^3 x^{3/2} dx + \int_3^9 (\frac{27}{2} - \frac{9}{2}x + x^{3/2}) dx$$

$$A = \left[ \frac{2}{5}x^{5/2} \right]_0^3 + \left[ \frac{27}{2}x - \frac{9}{4}x^2 + \frac{2}{5}x^{5/2} \right]_3^9$$

After calculating the definite integrals:

$$A = 16.2$$

$$10A = 162$$

Correct Option: B