Answer
${\bf{i}} \times {\bf{X}} = {\bf{v}}$ has a solution if and only if ${\bf{v}}$ is contained in the $yz$-plane, that is, the $i$-component is zero.
Work Step by Step
Let ${\bf{X}} = \left( {x,y,z} \right)$ and ${\bf{v}} = \left( {{v_1},{v_2},{v_3}} \right)$. We have ${\bf{i}} = \left( {1,0,0} \right)$, so
${\bf{i}} \times {\bf{X}} = {\bf{v}}$
(1) ${\ \ \ }$ $\left( {1,0,0} \right) \times \left( {x,y,z} \right) = \left( {{v_1},{v_2},{v_3}} \right)$
Consider the left-hand side of equation (1):
$\left( {1,0,0} \right) \times \left( {x,y,z} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&0\\
x&y&z
\end{array}} \right|$
$\left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&0\\
x&y&z
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
0&0\\
y&z
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
1&0\\
x&z
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
1&0\\
x&y
\end{array}} \right|{\bf{k}}$
So,
$\left( {1,0,0} \right) \times \left( {x,y,z} \right) = - z{\bf{j}} + y{\bf{k}}$
Now, the right-hand side of equation (1) is $\left( {{v_1},{v_2},{v_3}} \right) = {v_1}{\bf{i}} + {v_2}{\bf{j}} + {v_3}{\bf{k}}$. So, equation (1) becomes
$ - z{\bf{j}} + y{\bf{k}} = {v_1}{\bf{i}} + {v_2}{\bf{j}} + {v_3}{\bf{k}}$.
Since ${\bf{i}}$, ${\bf{j}}$ and ${\bf{k}}$ are linearly independent, the solution is
${\bf{v}} = \left( {0, - z,y} \right)$
This implies that ${\bf{v}}$ is contained in the $yz$-plane, that is, the $i$-component is zero.
Conversely, if ${\bf{v}}$ is contained in the $yz$-plane. We write ${\bf{v}} = \left( {0,{v_2},{v_3}} \right) = {v_2}{\bf{j}} + {v_3}{\bf{k}}$. Then equation (1) yields
$\left( {1,0,0} \right) \times \left( {x,y,z} \right) = - z{\bf{j}} + y{\bf{k}} = {v_2}{\bf{j}} + {v_3}{\bf{k}}$
So, it has solution ${\bf{v}} = \left( {0, - z,y} \right)$.
Hence, ${\bf{i}} \times {\bf{X}} = {\bf{v}}$ has a solution if and only if ${\bf{v}}$ is contained in the $yz$-plane, that is, the $i$-component is zero.
