To find the minimum value of the function f(x) = x^3 - 3x^2 - 24x + 100 in the interval [-3, 3], we can use calculus.

1. Find the derivative of the function:
f'(x) = 3x^2 - 6x - 24

2. Set the derivative equal to zero and solve for x:
3x^2 - 6x - 24 = 0

We can solve this quadratic equation by factoring or using the quadratic formula. Factoring gives us:
(x - 4)(3x + 6) = 0

So, x - 4 = 0 or 3x + 6 = 0
x = 4 or x = -2

3. Determine the second derivative of the function:
f''(x) = 6x - 6

4. Substitute the critical points (x = 4 and x = -2) into the second derivative to determine the nature of the points:
f''(4) = 6(4) - 6 = 18, which is positive
f''(-2) = 6(-2) - 6 = -18, which is negative

5. Analyze the results:
- When the second derivative is positive, it indicates that the function is concave up, and the critical point is a local minimum.
- When the second derivative is negative, it indicates that the function is concave down, and the critical point is a local maximum.

Since f''(4) is positive, x = 4 is a local minimum, and since f''(-2) is negative, x = -2 is a local maximum.

6. Evaluate the function at the endpoints of the interval:
f(-3) = (-3)^3 - 3(-3)^2 - 24(-3) + 100 = 37
f(3) = 3^3 - 3(3)^2 - 24(3) + 100 = -19

7. Compare the values at the critical points and endpoints:
f(-3) = 37
f(4) = 60
f(3) = -19

Since the local minimum at x = 4 has a lower value than the endpoints, the minimum value of the function in the interval [-3, 3] is 60. Therefore, the correct answer is option 'B' (28).