To solve the problem, follow these steps:

1. Find the gradient of \(L_1\) using points \(A(2, 5)\) and \(B(10, 9)\):

\(\text{Gradient of } L_1 = \frac{9 - 5}{10 - 2} = \frac{1}{2}\)

2. Since \(L_2\) is parallel to \(L_1\), its equation is:

\(y = \frac{1}{2}x\)

3. The line \(AC\) is perpendicular to \(L_2\), so its gradient is the negative reciprocal:

\(\text{Gradient of } AC = -2\)

4. Using point \(A(2, 5)\), the equation of line \(AC\) is:

\(y - 5 = -2(x - 2)\)

5. Solve the equations \(y = \frac{1}{2}x\) and \(y - 5 = -2(x - 2)\) simultaneously to find \(C\):

Substitute \(y = \frac{1}{2}x\) into \(y - 5 = -2(x - 2)\):

\(\frac{1}{2}x - 5 = -2(x - 2)\)

\(\frac{1}{2}x - 5 = -2x + 4\)

\(\frac{1}{2}x + 2x = 9\)

\(\frac{5}{2}x = 9\)

\(x = \frac{18}{5} = 3.6\)

\(y = \frac{1}{2}(3.6) = 1.8\)

Thus, \(C(3.6, 1.8)\).

6. Calculate the distance \(AC\):

\(AC = \sqrt{(3.6 - 2)^2 + (1.8 - 5)^2}\)

\(AC = \sqrt{1.6^2 + 3.2^2}\)

\(AC = \sqrt{2.56 + 10.24}\)

\(AC = \sqrt{12.8}\)

\(AC \approx 3.58\)