To calculate the area of triangle \( XYZ \) with vertices at points \( X(6, 8) \), \( Y(3, 3) \), and \( Z(13, -3) \), we can use the formula for the area of a triangle given by its vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of the points \( X(6, 8) \), \( Y(3, 3) \), and \( Z(13, -3) \):
- \( x_1 = 6, y_1 = 8 \)
- \( x_2 = 3, y_2 = 3 \)
- \( x_3 = 13, y_3 = -3 \)
Now we can plug these into the formula:
\[ \text{Area} = \frac{1}{2} \left| 6(3 - (-3)) + 3((-3) - 8) + 13(8 - 3) \right| \]
Calculating each term:
- \( 6(3 - (-3)) = 6(3 + 3) = 6 \times 6 = 36 \)
- \( 3((-3) - 8) = 3(-3 - 8) = 3(-11) = -33 \)
- \( 13(8 - 3) = 13(5) = 65 \)
Now substituting these results back into the area formula:
\[ \text{Area} = \frac{1}{2} \left| 36 - 33 + 65 \right| \]
Calculating inside the absolute value:
\[ 36 - 33 + 65 = 36 + 32 = 68 \]
Thus, we have:
\[ \text{Area} = \frac{1}{2} \left| 68 \right| = \frac{1}{2} \times 68 = 34 \]
Therefore, the area of triangle \( XYZ \) is:
\[ \boxed{34} \]