If ` z_1 ` lies on `|z-3| + |z + 3| = 8` such that arg `z_1 = pi//6 `, then ` 37|z_1 |^(2)` = _________.

Explanation

To solve the problem, we need to find \( 37 |z_1|^2 \) given that \( z_1 \) lies on the ellipse defined by \( |z - 3| + |z + 3| = 8 \) and \( \arg(z_1) = \frac{\pi}{6} \).

### Step-by-Step Solution:

1. **Understanding the Ellipse Equation**:
The equation \( |z - 3| + |z + 3| = 8 \) represents an ellipse with foci at \( z = 3 \) and \( z = -3 \). The total distance from any point on the ellipse to the two foci is constant and equals 8.

2. **Finding the Lengths**:
The distance between the foci is \( 6 \) (from \( -3 \) to \( 3 \)). Thus, the semi-major axis \( a \) can be found as follows:
\[ 2a = 8 \implies a = 4 \]

3. **Calculating the Eccentricity**:
The distance from the center of the ellipse to each focus \( c \) is \( 3 \) (since the foci are at \( -3 \) and \( 3 \)):
\[ c = 3 \]
The relationship between \( a \), \( b \), and \( c \) in an ellipse is given by:
\[ c^2 = a^2 - b^2 \]
Substituting the known values:
\[ 3^2 = 4^2 - b^2 \implies 9 = 16 - b^2 \implies b^2 = 7 \implies b = \sqrt{7} \]

4. **Parametric Representation of \( z_1 \)**:
Since \( \arg(z_1) = \frac{\pi}{6} \), we can express \( z_1 \) in terms of its modulus \( r \):
\[ z_1 = r \left( \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} \right) = r \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \]
Thus, we have:
\[ x_1 = r \cdot \frac{\sqrt{3}}{2}, \quad y_1 = r \cdot \frac{1}{2} \]

5. **Substituting into the Ellipse Equation**:
The standard form of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( x_1 \) and \( y_1 \):
\[ \frac{\left(r \cdot \frac{\sqrt{3}}{2}\right)^2}{16} + \frac{\left(r \cdot \frac{1}{2}\right)^2}{7} = 1 \]
Simplifying:
\[ \frac{r^2 \cdot \frac{3}{4}}{16} + \frac{r^2 \cdot \frac{1}{4}}{7} = 1 \]
This leads to:
\[ \frac{3r^2}{64} + \frac{r^2}{28} = 1 \]

6. **Finding a Common Denominator**:
The common denominator for \( 64 \) and \( 28 \) is \( 448 \):
\[ \frac{3r^2 \cdot 7}{448} + \frac{r^2 \cdot 16}{448} = 1 \]
Thus:
\[ \frac{21r^2 + 16r^2}{448} = 1 \implies \frac{37r^2}{448} = 1 \implies r^2 = \frac{448}{37} \]

7. **Calculating \( 37 |z_1|^2 \)**:
Now, we need to find \( 37 |z_1|^2 \):
\[ 37 |z_1|^2 = 37 \cdot r^2 = 37 \cdot \frac{448}{37} = 448 \]

### Final Answer:
\[ \boxed{448} \]