If the system of equations x - 3y + 5z = 3 x - 2y + 4z = 4 2x - 7y +…

The correct answer is Option (4) → 11

With infinitely many solutions the system cannot have a unique solution, so the determinant of the coefficients must be zero:

    \(\displaystyle \begin{vmatrix}1 & -3 & 5\\1 & -2 & 4\\2 & -7 & \lambda\end{vmatrix} = 0\)

Expanding along the first row:

    \(1(-2\lambda + 28) + 3(\lambda - 8) + 5(-7 + 4) = 0\)
    \(-2\lambda + 28 + 3\lambda - 24 - 15 = 0\)
    \(\lambda - 11 = 0,\quad \lambda = 11\)

A zero determinant allows either no solution or infinitely many, so the equations must also be consistent. For \(\lambda = 11\), the third equation is 3 times the first minus the second:

    \(3(x - 3y + 5z) - (x - 2y + 4z) = 2x - 7y + 11z\)
    \(3(3) - 4 = 5\)

So the third equation adds nothing new, and the system has infinitely many solutions when \(\lambda = 11\).