First, find the x-coordinate of the intersection point \(P\) by setting \(6 - 3x = 3\), which gives \(x = 1\).

Next, find the area under the curve from \(x = 0\) to \(x = 1\):

\(\int_0^1 \frac{9}{(5x+4)^{\frac{1}{2}}} \, dx = \left[ \frac{18}{5} (5x+4)^{\frac{1}{2}} \right]_0^1\)

Calculate the definite integral:

\(\frac{18}{5} \times 3 - \frac{18}{5} \times 2 = \frac{18}{5}\)

The line crosses the x-axis at \(x = 2\), forming a triangle with the x-axis and the line from \(x = 1\) to \(x = 2\). The area of this triangle is:

\(\frac{1}{2} \times 3 \times 1 = \frac{3}{2}\)

The total area of the shaded region is:

\(\frac{18}{5} + \frac{3}{2} = \frac{51}{10}\)