Using the formula

A × B = |A| |B| sin θ n

and our values from before which were

|BS| = √3
|CP| = √3
θ = 70.5°

we have:

|BS| × |CP|

= |BS| |CP| sin θ n = (√3)(√3) sin 70.5° n = 2.828 n

This means our result is a vector with magnitude 2.828 units and direction perpendicular to the plane containing BS and CP, indicated in green in the following diagram.