Answer
$\left( {1,1,1} \right) \times {\bf{X}} = \left( {1,0,0} \right)$ contradicts the fact that $\left( {1,0,0} \right)$ is not orthogonal to $\left( {1,1,1} \right)$. Hence, there is no solution.
Work Step by Step
Using the geometric description of the cross product given by Theorem 1, we have the vector $\left( {1,1,1} \right) \times {\bf{X}}$ orthogonal to $\left( {1,1,1} \right)$ and ${\bf{X}}$.
Since $\left( {1,1,1} \right) \times {\bf{X}} = \left( {1,0,0} \right)$, we must have $\left( {1,0,0} \right)$ orthogonal to $\left( {1,1,1} \right)$ and ${\bf{X}}$. However, the dot product is $\left( {1,0,0} \right)\cdot\left( {1,1,1} \right) = 1 \ne 0$. This implies that $\left( {1,0,0} \right)$ is not orthogonal to $\left( {1,1,1} \right)$. Thus, it is a contradiction. Hence, there is no solution.
