Answer
We have the solution $a = - 1$. Since the solution is unique, we conclude that the equation $\left( {1,2,3} \right) \times {\bf{v}} = \left( { - 1,2,a} \right)$ has no solution for $a \ne - 1$.
Work Step by Step
Let the components of ${\bf{v}}$ be ${\bf{v}} = \left( {{v_1},{v_2},{v_3}} \right)$. So,
$\left( {1,2,3} \right) \times {\bf{v}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&2&3\\
{{v_1}}&{{v_2}}&{{v_3}}
\end{array}} \right|$
$\left( {1,2,3} \right) \times {\bf{v}} = \left| {\begin{array}{*{20}{c}}
2&3\\
{{v_2}}&{{v_3}}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
1&3\\
{{v_1}}&{{v_3}}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
1&2\\
{{v_1}}&{{v_2}}
\end{array}} \right|{\bf{k}}$
$\left( {1,2,3} \right) \times {\bf{v}} = \left( {2{v_3} - 3{v_2}} \right){\bf{i}} - \left( {{v_3} - 3{v_1}} \right){\bf{j}} + \left( {{v_2} - 2{v_1}} \right){\bf{k}}$
Since $\left( {1,2,3} \right) \times {\bf{v}} = \left( { - 1,2,a} \right)$, so
$\left( {2{v_3} - 3{v_2}} \right){\bf{i}} - \left( {{v_3} - 3{v_1}} \right){\bf{j}} + \left( {{v_2} - 2{v_1}} \right){\bf{k}} = \left( { - 1,2,a} \right)$
We obtain a system of equations
$2{v_3} - 3{v_2} = - 1$
$ - {v_3} + 3{v_1} = 2$
${v_2} - 2{v_1} = a$
From the third equation we obtain ${v_1} = \frac{1}{2}\left( {{v_2} - a} \right)$. Substituting it in the second equation gives
$ - {v_3} + \frac{3}{2}\left( {{v_2} - a} \right) = 2$
$ - {v_3} + \frac{3}{2}{v_2} = 2 + \frac{3}{2}a$
Now, we solve two equations in terms of ${v_2}$ and ${v_3}$:
$2{v_3} - 3{v_2} = - 1$
$ - {v_3} + \frac{3}{2}{v_2} = 2 + \frac{3}{2}a$
We have from the second equation ${v_3} = \frac{3}{2}{v_2} - 2 - \frac{3}{2}a$. Substituting it in the first equations gives
$2\left( {\frac{3}{2}{v_2} - 2 - \frac{3}{2}a} \right) - 3{v_2} = - 1$
$3{v_2} - 4 - 3a - 3{v_2} = - 1$
$3a = - 3$, ${\ \ \ }$ $a = - 1$
We have $a = - 1$. Since the solution is unique, we conclude that the equation $\left( {1,2,3} \right) \times {\bf{v}} = \left( { - 1,2,a} \right)$ has no solution for $a \ne - 1$.
