JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Matrices

Dividing the third equation by $-2$ gives $x+y+2z=-\frac{3}{2}$. Adding this to $3x-y-2z=2$ gives $4x=\frac{1}{2}$, so $x=\frac{1}{8}$ in every solution. Subtracting $x+y+2z=-\frac{3}{2}$ from $kx+y+2z=1$ gives $(k-1)x=\frac{5}{2}$, so $(k-1)\frac{1}{8}=\frac{5}{2}$ and $k=21$. With this value the system reduces to $x=\frac{1}{8}$ together with $y+2z=-\frac{13}{8}$, which has infinitely many solutions. The required value is $k=21$.