The system of equations
$x + y + z = 7$
$x + 2y + 3z = 5$
$x + 3y+λz = μ$
has a unique solution, if

λ ≠ 5 and μ is any real number

The correct answer is Option (3) → λ ≠ 5 and μ is any real number

Given system:

$\begin{aligned} (1)\quad & x + y + z = 7 \\ (2)\quad & x + 2y + 3z = 5 \\ (3)\quad & x + 3y + \lambda z = \mu \end{aligned}$

Write as augmented matrix:

$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 7 \\ 1 & 2 & 3 & 5 \\ 1 & 3 & \lambda & \mu \end{array}\right]$

Apply $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:

$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 7 \\ 0 & 1 & 2 & -2 \\ 0 & 2 & \lambda - 1 & \mu - 7 \end{array}\right]$

Next, $R_3 \rightarrow R_3 - 2R_2$:

$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 7 \\ 0 & 1 & 2 & -2 \\ 0 & 0 & \lambda - 5 & \mu - 3 \end{array}\right]$

Unique solution exists if the coefficient matrix has full rank (3) ⟹

The third pivot $\lambda - 5 \ne 0$

Condition for unique solution: $\lambda \ne 5$