Answer
(a) $-6$
(b) No, 6.
Work Step by Step
(a) Given the three vectors, first calculate the cross product of the two vectors: $\vec v\times\vec w
=(-1\times0+1\times0)i+ (-1\times6-3\times0)j+ (3\times0+1\times6)k =0i-6 j +6k$. Next we calculate the scalar triple product as $\vec u\cdot(\vec v\times\vec w)=2\times0-2\times(-6)-3\times6=-6$
(b) As the scalar triple product is not zero, we conclude that these vectors are not coplanar, and the volume of the parallelepiped is 6.
