Answer
The solution is $\left( {x,y,z} \right) = \left( {k,k,1 + k} \right)$, where $k$ is any number. This implies that there are infinitely many solutions.
Work Step by Step
Since ${\bf{X}} = \left( {x,y,z} \right)$ and $\left( {1, - 1,0} \right) = {\bf{i}} - {\bf{j}}$, so
$\left( {1,1,1} \right) \times {\bf{X}} = \left( {1, - 1,0} \right)$
$\left( {1,1,1} \right) \times \left( {x,y,z} \right) = {\bf{i}} - {\bf{j}}$
Let us consider the left-hand side of the equation. Using the formula for the cross product, we have
$\left( {1,1,1} \right) \times \left( {x,y,z} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&1&1\\
x&y&z
\end{array}} \right|$
Since
$\left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&1&1\\
x&y&z
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
1&1\\
y&z
\end{array}} \right){\bf{i}} - \left( {\begin{array}{*{20}{c}}
1&1\\
x&z
\end{array}} \right){\bf{j}} + \left( {\begin{array}{*{20}{c}}
1&1\\
x&y
\end{array}} \right){\bf{k}}$
So,
$\left( {1,1,1} \right) \times \left( {x,y,z} \right) = \left( {z - y} \right){\bf{i}} - \left( {z - x} \right){\bf{j}} + \left( {y - x} \right){\bf{k}}$
and
$\left( {z - y} \right){\bf{i}} - \left( {z - x} \right){\bf{j}} + \left( {y - x} \right){\bf{k}} = {\bf{i}} - {\bf{j}}$
Thus, we obtain a system of equations:
$z-y=1$, ${\ \ }$ $z-x=1$, ${\ \ }$ $y-x=0$.
The third equation above gives $x=y$.
If we write $k=x=y$. then the solution is $\left( {x,y,z} \right) = \left( {k,k,1 + k} \right)$, where $k$ is any number. This implies that there are infinitely many solutions.
