JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For the given system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, = 0. = vmatrix 1 & 5 & 6 \\ 2 & 3 & 4 \\ 1 & 6 & a vmatrix = 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 a = 50 7 For infinitely many solutions, we must also have _z = 0. _z = vmatrix 1 & 5 & 4 \\ 2 & 3 & 7 \\ 1 & 6 & b vmatrix = 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 b = 29 7 The point (a, b) is ( 50 7 , 29 7 ). Checking the given options, we evaluate a - b: a - b = 50 7 - 29 7 = 21 7 = 3 Therefore, the point (a, b) lies on the line x - y = 3. Answer: x - y = 3