Chapter 5 - Inner Product Spaces - 5.5 Applications of Inner Product Spaces - 5.5 Exercises - Page 282: 48

$6$

We know that $ai+bj+ck=(a,b,c).$ We know that for a matrix $ \left[\begin{array}{rrr} a & b & c \\ d &e & f \\ g &h & i \\ \end{array} \right] $ the determinant, $D=a(ei-fh)-b(di-fg)+c(dh-eg).$ $v×w$ is the determinant of the matrix $\begin{bmatrix} i& j & k \\ 0& 3&0\\ 0&0 &1 \\ \end{bmatrix} $ Thus $v×w=(3,0,0).$ Thus $u\cdot (v\times w)=(2,0,1)(3,0,0)=6+0+0=6$