JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Matrices

A non-trivial solution exists only when the coefficient determinant vanishes: $\left|\begin{array}{ccc}2& 3& -1\\ 1& k& -2\\ 2& -1& 1\end{array}\right|=0$, which expands to $2(k-2)-3(1+4)-1(-1-2k)=0$ and gives $k=\frac{9}{2}$. Taking the equations, subtracting the third from the first gives $4y-2z=0$, so $z=2y$ and $\frac{y}{z}=\frac{1}{2}$. Putting $z=2y$ into the first equation, $2x+3y-2y=0$ gives $x=-\frac{y}{2}$, so $\frac{x}{y}=-\frac{1}{2}$ and $\frac{z}{x}=-4$. Adding everything, $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k=-\frac{1}{2}+\frac{1}{2}-4+\frac{9}{2}=\frac{1}{2}$.