JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For the given system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. vmatrix 3 & -8 & -12 \\ 2 & 3 & 3 \\ 1 & 1 & 3 vmatrix = 0 Expanding the determinant along the first row: 3 (9 - 3) - (-8)(3 2 - 3) + (-12)( 2 - 3) = 0 6 3 + 24 2 - 24 - 12 2 + 36 = 0 6 3 + 12 2 + 12 = 0 Dividing by 6, we get: 3 + 2 2 + 2 = 0 Using the multiple angle formulas 3 = 4 ^3 - 3 and 2 = 2 ^2 - 1, we substitute these into the equation: (4 ^3 - 3 ) + 2(2 ^2 - 1) + 2 = 0 4 ^3 + 4 ^2 - 3 = 0 (4 ^2 + 4 - 3) = 0 (2 - 1)(2 + 3) = 0 This gives the possible values for : = 0 = 2 , 3 2 = 1 2 = 3 , 5 3 = - 3 2 (Not possible since [-1, 1]) The possible values of [0, 2 ] are 2 , 3 2 , 3 , 5 3 . Sum of all possible values of : = ( 2 + 3 2 ) + ( 3 + 5 3 ) = 2 + 2 = 4 Answer: 4