Find the area of the parallelogram that has adjacent sides u = i-3j+2k and v = 3j-k

The area is the given by the magnitude of the cross product of the side vectors. That value is

Area = ||u*v|| = ||u||·||v||·sin (θ)

where θ is the angle between the vectors.

The cross product of [1, - 3, 2] and [0, 3, - 1] is [-3, 1, 3}, which has magnitude

√ ((-3) ² + 1² + 3²) = √19 ≈ 4.3589