To solve the equation \(|x-3| + |3x-6| - |5-2x| = 8\), we need to consider the intervals determined by the critical points where the expressions inside the absolute values change sign. These critical points are \(x = 3\), \(x = 2\), and \(x = \frac{5}{2}\).

Let's analyze the equation on the following intervals:

1. **Interval 1: \(x < 2\)**
- Here, \(x - 3 < 0\), so \(|x-3| = -(x-3) = -x + 3\).
- \(3x - 6 < 0\), so \(|3x-6| = -(3x-6) = -3x + 6\).
- \(5 - 2x > 0\), so \(|5-2x| = 5 - 2x\).

Substituting these into the equation gives:
\[
(-x + 3) + (-3x + 6) - (5 - 2x) = -x + 3 - 3x + 6 - 5 + 2x = -2x + 4 = 8
\]
Solving \(-2x + 4 = 8\):
\[
-2x = 4 \implies x = -2
\]
Since \(-2 < 2\), \(x = -2\) is valid in this interval.

2. **Interval 2: \(2 \le x < \frac{5}{2}\)**
- Here, \(x - 3 < 0\), so \(|x-3| = -(x-3) = -x + 3\).
- \(3x - 6 < 0\), so \(|3x-6| = -(3x-6) = -3x + 6\).
- \(5 - 2x > 0\), so \(|5-2x| = 5 - 2x\).

Substituting these into the equation gives:
\[
(-x + 3) + (-3x + 6) - (5 - 2x) = -x + 3 - 3x + 6 - 5 + 2x = -2x + 4 = 8
\]
Solving \(-2x + 4 = 8\):
\[
-2x = 4 \implies x = -2
\]
However, \(x = -2\) is not valid in this interval because \(-2 < 2\).

3. **Interval 3: \(\frac{5}{2} \le x < 3\)**

- Here, \(x - 3 < 0\), so \(|x-3| = -(x-3) = -x + 3\).
- \(3x - 6 < 0\), so \(|3x-6| = -(3x-6) = -3x + 6\).
- \(5 - 2x < 0\), so \(|5-2x| = -(5-2x) = 2x - 5\).

Substituting these into the equation gives:
\[
(-x + 3) + (-3x + 6) - (2x - 5) = -x + 3 - 3x + 6 - 2x + 5 = -6x + 14 = 8
\]
Solving \(-6x + 14 = 8\):
\[
-6x = -6 \implies x = 1
\]
However, \(x = 1\) is not valid in this interval because \(x = 1 < \frac{5}{2}\).

4. **Interval 4: \(x \ge 3\)**
- Here, \(x - 3 \ge 0\), so \(|x-3| = x - 3\).
- \(3x - 6 \ge 0\), so \(|3x-6| = 3x - 6\).
- \(5 - 2x < 0\), so \(|5-2x| = -(5-2x) = 2x - 5\).

Substituting these into the equation gives:
\[
(x - 3) + (3x - 6) - (2x - 5) = x - 3 + 3x - 6 - 2x + 5 = 2x - 4 = 8
\]
Solving \(2x - 4 = 8\):
\[
2x = 12 \implies x = 6
\]
Since \(x = 6 \ge 3\), \(x = 6\) is valid in this interval.

From these intervals, the solutions to the original equation are \(x = -2\) and \(x = 6\).