Triangle ABC has vertices at A(2,1,2), B(3,0,2) and C(-1,2,1) respectively. Find Angle BAC.

I presume the question is asking to work out Angle A, but I am a bit confused. I know that you are suppose to use the dot product and formula:

$\displaystyle cos\theta = \frac {\mathbf{a}\cdot \mathbf{b}}{|\boldsymbol{a}||\mathbf{b}|}$

Are you suppose to use $\displaystyle \mathbf{b}\cdot \mathbf{c}$ since Angles A lies between side B and C?