JEE Main 2026 — Determinants Question with Solution

For the given system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, $\Delta =0$.

$\Delta =\left|\begin{array}{ccc}1& 5& 6\\ 2& 3& 4\\ 1& 6& a\end{array}\right|=0$

Expanding along the first row:

$1(3a-24)-5(2a-4)+6(12-3)=0$

$3a-24-10a+20+54=0$

$-7a+50=0\Rightarrow a=\frac{50}{7}$

For infinitely many solutions, we must also have ${\Delta }_z=0$.

${\Delta }_z=\left|\begin{array}{ccc}1& 5& 4\\ 2& 3& 7\\ 1& 6& b\end{array}\right|=0$

Expanding along the first row:

$1(3b-42)-5(2b-7)+4(12-3)=0$

$3b-42-10b+35+36=0$

$-7b+29=0\Rightarrow b=\frac{29}{7}$

The point $(a,b)$ is $\left(\frac{50}{7},\frac{29}{7}\right)$.

Checking the given options, we evaluate $a-b$:

$a-b=\frac{50}{7}-\frac{29}{7}=\frac{21}{7}=3$

Therefore, the point $(a,b)$ lies on the line $x-y=3$.

Answer: $x-y=3$