JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Matrices

The system is homogeneous, so non-trivial solutions exist precisely when the coefficient determinant vanishes. Expanding $\left|\begin{array}{ccc}7& 6& -2\\ 3& 4& 2\\ 1& -2& -6\end{array}\right|=7(-24+4)-6(-18-2)-2(-6-4)=-140+120+20=0$ confirms this, so there are infinitely many solutions. Adding the first two equations gives $10x+10y=0$, so $y=-x$; substituting into $x-2y-6z=0$ gives $3x=6z$, that is $x=2z$. The solutions are infinitely many, satisfying $x=2z$.