If the equation `x^(2)-3x+|3x-a|le0` is to be satisfied by atleast one `xgt0`, then let Mbe the interval of values of a needed, then integral part of length of M is -----------

Answer

To solve the inequality \( x^2 - 3x + |3x - a| \leq 0 \) for at least one \( x > 0 \), we will analyze the cases based on the expression inside the absolute value.

### Step 1: Analyze the cases for \( |3x - a| \)

The absolute value \( |3x - a| \) can be split into two cases:

1. **Case 1:** \( 3x - a \geq 0 \) (i.e., \( x \geq \frac{a}{3} \))
- In this case, \( |3x - a| = 3x - a \).
- The inequality becomes:
\[ x^2 - 3x + (3x - a) \leq 0 \implies x^2 - a \leq 0 \]
- This simplifies to:
\[ x^2 \leq a \]
- Thus, we have \( 0 < x \leq \sqrt{a} \).

2. **Case 2:** \( 3x - a < 0 \) (i.e., \( x < \frac{a}{3} \))
- In this case, \( |3x - a| = a - 3x \).
- The inequality becomes:
\[ x^2 - 3x + (a - 3x) \leq 0 \implies x^2 - 6x + a \leq 0 \]
- This is a quadratic inequality in \( x \).

### Step 2: Find the roots of the quadratic \( x^2 - 6x + a = 0 \)

The roots of the quadratic equation \( x^2 - 6x + a = 0 \) are given by:
\[ x = \frac{6 \pm \sqrt{36 - 4a}}{2} = 3 \pm \sqrt{9 - a} \]
Let the roots be \( r_1 = 3 - \sqrt{9 - a} \) and \( r_2 = 3 + \sqrt{9 - a} \).

### Step 3: Determine the conditions for \( x \) to be in the interval

For the quadratic \( x^2 - 6x + a \leq 0 \) to have real roots, the discriminant must be non-negative:
\[ 36 - 4a \geq 0 \implies a \leq 9 \]

### Step 4: Find the intersection of intervals

For the inequality \( x^2 \leq a \) to hold, we need:
- \( x \) must be in the interval \( \left(0, \sqrt{a}\right) \).
- For the second case, \( x \) must be in the interval \( \left(r_1, r_2\right) \).

To ensure that there is at least one \( x > 0 \) that satisfies both conditions, we need:
\[ \frac{a}{3} < r_2 \quad \text{and} \quad r_1 < \sqrt{a} \]

### Step 5: Solve the inequalities

1. From \( \frac{a}{3} < r_2 \):
\[ \frac{a}{3} < 3 + \sqrt{9 - a} \implies a < 9 + 3\sqrt{9 - a} \]
This inequality will hold for \( a \) in the range \( (0, 9) \).

2. From \( r_1 < \sqrt{a} \):
\[ 3 - \sqrt{9 - a} < \sqrt{a} \implies 3 < \sqrt{a} + \sqrt{9 - a} \]
Squaring both sides gives:
\[ 9 < a + 9 - a + 2\sqrt{a(9 - a)} \implies 0 < 2\sqrt{a(9 - a)} \implies a(9 - a) > 0 \]
This is satisfied when \( 0 < a < 9 \).

### Step 6: Conclusion

Thus, the interval \( M \) of values for \( a \) is \( (0, 9) \). The length of this interval \( M \) is \( 9 - 0 = 9 \).

### Final Answer

The integral part of the length of \( M \) is \( \boxed{9} \).