If someone could tell me how on mathematica you do this thanks?

FInd the exact expression for the Sine of the angle between vectors <2,-1,1> and <3,2,1>.

Calculate the cross product with Cross and go from there.

Now stuff has become so easy …

In[3]:= Sin[VectorAngle[{2, -1, 1}, {3, 2, 1}]]
Out[3]= Sqrt[59/21]/2

VectorAngle and the formula is of course

In[4]:= Sin[ArcCos[Dot[{2, -1, 1}, {3, 2, 1}]/(Norm[{2, -1, 1}] Norm[{3, 2, 1}])]]
Out[4]= Sqrt[59/21]/2

as the help spells it out.