If $A$ = $(a, 0, 0)$; $B$ = $(0,b,0)$ & $C$ = $(0,0,c)$: Find (A $\times$ B) $\cdotp$ C. I know that
$A \times B = (A_yB_z - A_zB_y)\hat{\imath}$ $  (A_zB_x - A_xB_z)\hat{\jmath} A_xB_y-A_yB_x)\hat{k}$

This would be $(0 \cdot 0 - 0\cdot b)\hat{\imath} (0\cdot 0 - a\cdot 0)\hat{\jmath} (a\cdot b - 0 \cdot 0)\hat{k}$
Which is just $0\hat{\imath} 0\hat{\jmath}(ab) \hat{k}$ right?
And thus $(A \times B)  \bullet C = ((AB)_xC_x+(AB)_yC_y+(AB)_zC_z).$ 
Which is $(0)_x+(0)_y+(ab\cdot c)_z = abc.$
If this is correct, could someone thoroughly explain the geometrical meaning of this?

$|(A \times B) \cdot C| =|abc|$ is the volume of the box which has as thee of the sides sides the vectors $A$, $B$ and $C$. This remains true for any three vectors $A$, $B$ and $C$, where in the general case the box may be tilted or squeezed, as $A$, $B$ and $C$ may not be perpendicular to each other.
