If `|z_1|=1, |z_2|=2, |z_3|=3` and `|9z_1z_2 + 4z_1z_3 + z_2z_3|=36`, then `|z_1+z_2+z_3|` is equal to _______.

Answer

To solve the problem, we need to find the value of \( |z_1 + z_2 + z_3| \) given the conditions \( |z_1| = 1 \), \( |z_2| = 2 \), \( |z_3| = 3 \), and \( |9z_1z_2 + 4z_1z_3 + z_2z_3| = 36 \).

### Step-by-Step Solution:

1. **Understanding the given moduli:**
- We have \( |z_1| = 1 \), \( |z_2| = 2 \), and \( |z_3| = 3 \).

2. **Expressing the modulus of the given expression:**
- We need to evaluate \( |9z_1z_2 + 4z_1z_3 + z_2z_3| \).
- We know that \( |z_1 z_2| = |z_1| |z_2| = 1 \cdot 2 = 2 \).
- Similarly, \( |z_1 z_3| = |z_1| |z_3| = 1 \cdot 3 = 3 \).
- And \( |z_2 z_3| = |z_2| |z_3| = 2 \cdot 3 = 6 \).

3. **Finding the modulus of each term:**
- Thus, we have:
- \( |9z_1 z_2| = 9 |z_1 z_2| = 9 \cdot 2 = 18 \)
- \( |4z_1 z_3| = 4 |z_1 z_3| = 4 \cdot 3 = 12 \)
- \( |z_2 z_3| = 6 \)

4. **Applying the triangle inequality:**
- By the triangle inequality, we can say:
\[ |9z_1 z_2 + 4z_1 z_3 + z_2 z_3| \leq |9z_1 z_2| + |4z_1 z_3| + |z_2 z_3| = 18 + 12 + 6 = 36 \]
- Since we are given that \( |9z_1 z_2 + 4z_1 z_3 + z_2 z_3| = 36 \), this means that the equality holds, which implies that the vectors \( 9z_1 z_2 \), \( 4z_1 z_3 \), and \( z_2 z_3 \) are collinear.

5. **Finding the modulus of the sum:**
- We can express the equality as:
\[ |9z_1 z_2 + 4z_1 z_3 + z_2 z_3| = |z_1 z_2 z_3| \cdot |9 \frac{z_1 z_2}{z_1 z_2 z_3} + 4 \frac{z_1 z_3}{z_1 z_2 z_3} + \frac{z_2 z_3}{z_1 z_2 z_3}| \]
- This simplifies to:
\[ |z_1 z_2 z_3| = |z_1| |z_2| |z_3| = 1 \cdot 2 \cdot 3 = 6 \]

6. **Final calculation:**
- Therefore, we have:
\[ |9z_1 z_2 + 4z_1 z_3 + z_2 z_3| = |z_1 z_2 z_3| \cdot |9 + 4 + 1| = 6 \cdot |z_1 + z_2 + z_3| \]
- Setting this equal to 36 gives:
\[ 6 |z_1 + z_2 + z_3| = 36 \]
- Thus, we find:
\[ |z_1 + z_2 + z_3| = \frac{36}{6} = 6 \]

### Conclusion:
The value of \( |z_1 + z_2 + z_3| \) is \( 6 \).