[THOUGHT] To find the area of triangle $ABC$, I can use the formula involving two sides and the included angle: [ \text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle BAC) ]

First, I need to determine $\sin(\angle BAC)$. Given that $\cos(\angle BAC) = -\frac{3}{5}$, and knowing that $\sin^2(\theta) + \cos^2(\theta) = 1$, I can find $\sin(\angle BAC)$.

Let's compute $\sin(\angle BAC)$. [/THOUGHT] [PYTHON] import math

Given values

cos_theta = -3/5

Calculate sin_theta

sin_theta = math.sqrt(1 - cos_theta**2) print(sin_theta)

[THOUGHT] Now that I have $\sin(\angle BAC) = 0.8$, I can compute the area of triangle $ABC$ using the formula: [ \text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle BAC) = \frac{1}{2} \times 5 \times 6 \times 0.8 ] Let's calculate this final value. [/THOUGHT] [PYTHON]

Calculate the area

AB = 5 AC = 6 sin_theta = 0.8 area = 0.5 * AB * AC * sin_theta print(area)

[THOUGHT] The area of triangle $ABC$ is calculated to be 12.0. Therefore, the final answer is: [/THOUGHT] \boxed{12}