Okay, so I need to find the angle ABC where the points are given in 3D space. Let me recall how to compute angles in 3D geometry. Hmm, angle ABC is the angle at point B between points A, B, and C. So, the vertex is B, and the two sides are BA and BC. To find this angle, I think I can use the dot product formula. The formula for the angle between two vectors is θ = arccos( (u · v) / (|u||v|) ). 

First, I need to find the vectors BA and BC. Wait, BA is from B to A, right? So, to get the vector BA, I subtract B's coordinates from A's coordinates. Similarly, vector BC is C's coordinates minus B's coordinates. Let me write that down.

Coordinates of A: (1, -11, 2)
Coordinates of B: (3, -4, 1)
Coordinates of C: (-2, 1, -1)

Vector BA = A - B = (1 - 3, -11 - (-4), 2 - 1) = (-2, -7, 1)
Vector BC = C - B = (-2 - 3, 1 - (-4), -1 - 1) = (-5, 5, -2)

Okay, so BA is (-2, -7, 1) and BC is (-5, 5, -2). Now, I need to compute the dot product of BA and BC. The dot product is calculated by multiplying the corresponding components and then adding them up.

Dot product BA · BC = (-2)*(-5) + (-7)*(5) + (1)*(-2) = 10 - 35 - 2 = (10 - 35) is -25, then -25 -2 is -27. So the dot product is -27.

Next, I need the magnitudes of BA and BC. The magnitude of a vector is the square root of the sum of the squares of its components.

Magnitude of BA: sqrt[ (-2)^2 + (-7)^2 + (1)^2 ] = sqrt[4 + 49 + 1] = sqrt[54] = 3*sqrt(6). Let me check that: 4 + 49 is 53, plus 1 is 54. Yes, sqrt(54) simplifies to 3*sqrt(6).

Magnitude of BC: sqrt[ (-5)^2 + (5)^2 + (-2)^2 ] = sqrt[25 + 25 + 4] = sqrt[54] = 3*sqrt(6) as well. Wait, same magnitude? Interesting. So both vectors BA and BC have the same length here.

So the cosine of the angle is (dot product)/(product of magnitudes) = (-27)/(3*sqrt(6)*3*sqrt(6)) = (-27)/(9*6) = (-27)/54 = -0.5.

So cos(theta) = -0.5. Then theta is arccos(-0.5). The arccos of -0.5 is 120 degrees. Since cosine is negative, the angle is in the second quadrant, which makes sense for an angle between two vectors in 3D space.

Wait, let me double-check all steps to make sure I didn't make a mistake.

First, vectors BA and BC:

BA: A - B = (1-3, -11 - (-4), 2 -1) = (-2, -7, 1). Correct.

BC: C - B = (-2 -3, 1 - (-4), -1 -1) = (-5, 5, -2). Correct.

Dot product: (-2)(-5) + (-7)(5) + (1)(-2) = 10 -35 -2 = -27. Correct.

Magnitudes:

|BA| = sqrt(4 + 49 +1) = sqrt(54). Correct.

|BC| = sqrt(25 +25 +4) = sqrt(54). Correct.

Dot product divided by product of magnitudes: (-27)/(sqrt(54)*sqrt(54)) = (-27)/54 = -0.5. Correct.

Arccos(-0.5) is 120 degrees. That seems right. Because cos(120°) = -0.5. Let me confirm with the unit circle. Yes, 120 degrees is in the second quadrant, and cos(120°) is indeed -0.5.

So, the angle ABC is 120 degrees. Hmm, that seems straightforward. Let me just visualize the vectors to make sure. Both BA and BC vectors have the same magnitude and the angle between them is 120 degrees. It's possible in 3D space for the angle to be greater than 90 degrees even if the vectors are not in the same plane. Wait, but in this case, since we computed the angle using the dot product formula, which accounts for the 3D nature, it's correct. 

Alternatively, if I think of the coordinates, point B is (3,-4,1). Point A is (1,-11,2), so moving from B to A, it's like going back in x, down a lot in y, and up in z. Point C is (-2,1,-1), so moving from B to C is going left in x, up in y, and down in z. The vectors BA and BC are in different directions, so the angle between them is indeed 120 degrees. 

I don't see any mistakes in the calculations. So the final answer should be 120 degrees.
Final Answer
\boxed{120^\circ}
</think>