Let `lambda` be a real number for which the system of linear equations
x + y +z =6, 4x + `lambday -lambdaz = lambda -2` and 3x + 2y-4z =-5
has infinitely many solutions. Then `lambda` is a root of the quadratic equation

Options

A. `lambda^(2)-3lambda-4 = 0`
B. `lambda^(2) + 3lambda-4 =0`
C. `lambda^(2)-lambda-6=0`
D. `lambda^(2) + lambda-6 =0`

Explanation

To find the value of \( \lambda \) for which the given system of linear equations has infinitely many solutions, we need to analyze the determinant of the coefficient matrix. If the determinant is zero, then the system has either no solutions or infinitely many solutions.

Let's denote the equations as follows:

1. \( x + y + z = 6 \) (Equation 1)
2. \( 4x + \lambda y - \lambda z = \lambda - 2 \) (Equation 2)
3. \( 3x + 2y - 4z = -5 \) (Equation 3)

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