[THOUGHT] To find the area of triangle ABC given sides AB = 5, AC = 6, and cos(∠BAC) = -3/5, we can use the formula for the area of a triangle involving two sides and the sine of the included angle. First, recall that sin²θ + cos²θ = 1 to find sin(∠BAC). Then multiply half the product of AB and AC by this sine value.]

[PYTHON] import math

Given values

AB = 5 AC = 6 cos_angle = -3/5

Calculate sin using Pythagorean identity: sin²θ + cos²θ = 1 → |sin θ|=sqrt(1-cos²θ)

sin_angle = math.sqrt(1 - (cos_angle ** 2)) # We take positive root since angle is between sides in a triangle and sine must be nonnegative here? Wait, actually cosine negative means obtuse so sin still nonnegative! Yes because range of angles: [0°-180°] area = 0.5 * AB * AC * sin_angle # Area formula using two sides & included angle's sine print(round(area)) # Output rounded to integer as per problem request format (though exact value is fractional)"

[THOUGHT] The area of triangle ABC is 12 based on accurate calculations using given side lengths and angle cosine value.] $\boxed{12}$