The system of equations

$x + y + z = 4$

$x + 2y + 3z = 12$

$x + 3y + \lambda z = \mu$ has a unique solution if

A system of linear equations has a unique solution when the coefficient matrix has a non-zero determinant.

For the system:

$x + y + z = 4$

$x + 2y + 3z = 12$

$x + 3y + \lambda z = \mu$

The coefficient matrix is:

$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{pmatrix}$

---

Using the first row expansion:

$|A| = 1 \times \begin{vmatrix} 2 & 3 \\ 3 & \lambda \end{vmatrix} - 1 \times \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} + 1 \times \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}$

Calculating each 2×2 determinant:

$\begin{vmatrix} 2 & 3 \\ 3 & \lambda \end{vmatrix} = 2\lambda - 9$

$\begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} = \lambda - 3$

$\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 3 - 2 = 1$

---

$|A| = 1(2\lambda - 9) - 1(\lambda - 3) + 1(1)$

$|A| = 2\lambda - 9 - \lambda + 3 + 1$

$|A| = \lambda - 5$

---

For a unique solution, the determinant must be non-zero:

$\lambda - 5 \neq 0$

$\lambda \neq 5$

The parameter $\mu$ appears only on the right-hand side of the equations and does not affect the determinant. Therefore, $\mu$ can be any real number.

---

The system has a unique solution if $\lambda \neq 5$ and $\mu$ can be any real number.