Answer
Choose:
${\bf{a}} = \left( {3,0,1} \right)$
${\bf{b}} = \left( {1,1,1} \right)$
${\bf{c}} = \left( { - 2,1,0} \right)$
${\bf{a}} \times {\bf{b}} = {\bf{a}} \times {\bf{c}} \ne {\bf{0}}$ but ${\bf{b}} \ne {\bf{c}}$.
Work Step by Step
We have the conditions: ${\bf{a}} \times {\bf{b}} = {\bf{a}} \times {\bf{c}} \ne {\bf{0}}$ but ${\bf{b}} \ne {\bf{c}}$.
Write
${\bf{a}} \times {\bf{b}} - {\bf{a}} \times {\bf{c}} = {\bf{0}}$
${\bf{a}} \times \left( {{\bf{b}} - {\bf{c}}} \right) = {\bf{0}}$
This implies that the vector ${\bf{b}} - {\bf{c}}$ is parallel to ${\bf{a}}$. We may choose ${\bf{a}} = {\bf{b}} - {\bf{c}}$.
For instance: ${\bf{b}} = \left( {1,1,1} \right)$, ${\bf{c}} = \left( { - 2,1,0} \right)$, so
${\bf{a}} = {\bf{b}} - {\bf{c}} = \left( {1,1,1} \right) - \left( { - 2,1,0} \right) = \left( {3,0,1} \right)$
Let us verify this:
${\bf{a}} \times {\bf{b}} = \left( {3,0,1} \right) \times \left( {1,1,1} \right) = \left( { - 1, - 2,3} \right)$
${\bf{a}} \times {\bf{c}} = \left( {3,0,1} \right) \times \left( { - 2,1,0} \right) = \left( { - 1, - 2,3} \right)$
Thus, ${\bf{a}} \times {\bf{b}} = {\bf{a}} \times {\bf{c}} \ne {\bf{0}}$ but ${\bf{b}} \ne {\bf{c}}$.
