Let $a,b,c$ be real numbers. I have got a system of equations: $$\begin{eqnarray} 2x_2y_3-2x_3y_2&=&a\\ 2x_3y_1-2x_1y_3&=&b\\ 2x_1y_2-2x_2y_1&=&c. \end{eqnarray}$$
If we choose $x_1,x_2,x_3$ first, then we will solve system of linear equations: $$\begin{pmatrix} 0&-2x_3&2x_2\\ 2x_3&0&-2x_1\\ -2x_2&2x_1&0 \end{pmatrix}\begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix}=\begin{pmatrix} a\\ b\\ c \end{pmatrix}.$$ However, the determinant of $\begin{pmatrix} 0&-2x_3&2x_2\\ 2x_3&0&-2x_1\\ -2x_2&2x_1&0 \end{pmatrix}$ is $0$. So, we are not sure to choose $y_1,y_2,y_3$. Is there any way to find a nonzero solution (i.e $x_1,x_2,x_3,y_1,y_2,y_3$ are not all $0$) to this system without solving for the variables explicitly?
Any counterexample or reference or technique is very much appreciated. I appreciate any help you can provide.