Well, vector AB will be:

(2 - 1)i + (3 - 2)j + (-1 - 3)k = i + j - 4k

Vector AC will be:

(3 - 1)i + (-1 - 2)j + (1 - 3)k = 2i - 3j - 2k

(where i, j, and k are the unit vectors in the x, y, and z directions, respectively.)

We also know that

(AB) (dot) (AC) = |AB|*|AC|* cos(theta)

where theta is the angle between the vectors.

(AB) (dot) (AC) = (1)*(2) + (1)*(-3) + (-4)*(-2) = 7

|AB| = sqrt{1^2 + 1^2 + 4^2} = sqrt{18} = 3*sqrt{2}

|AC| = sqrt{2^2 + 3^2 + 2^2} = sqrt{17}

So cos(theta) = 7/(3*sqrt{2}*sqrt{17})

Now take the inverse cosine to get the angle.

-Dan