Answer
See below.
Work Step by Step
Let $u=(a,b,c)$.
We know that for a matrix
$
\left[\begin{array}{rrr}
a & b & c \\
d &e & f \\
g &h & i \\
\end{array} \right]
$
the determinant, $D=a(ei-fh)-b(di-fg)+c(dh-eg).$
$u×0$ is the determinant of the matrix $\begin{bmatrix}
i& j & k \\
a& b&c\\
0&0 &0 \\
\end{bmatrix}
$
Thus $u×0=(b\cdot0-c\cdot0,c\cdot0-a\cdot0,a\cdot0-b\cdot0)=(0,0,0).$
We also know that $0\times u=-u\times 0=-(0,0,0)=(0,0,0)$.
Thus, we proved what we had to.