The system of equations

$x + y + z = 7$

$x + 2y + 3z = 5$

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

has a unique solution, if

For a system of equations to have a unique solution, the determinant of the coefficient matrix must be non-zero.

Why? When the determinant is zero, the equations are dependent on each other, giving infinite solutions or no solution. When it's non-zero, the equations are independent and give exactly one answer.

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The coefficient matrix contains the numbers in front of $x$, $y$, and $z$:

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

Note: We ignore the right-hand side $(7, 5, \mu)$ because we only need the coefficients to check uniqueness.

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Using first row expansion (the easiest method for $3 \times 3$ matrices):

$\det(A) = 1 \begin{vmatrix} 2 & 3 \\ 3 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}$

Computing each $2 \times 2$ determinant using $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$:

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

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

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

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Combining everything:

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

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

$= \lambda - 5$

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For a unique solution:

$\det(A) \neq 0$

$\lambda - 5 \neq 0$

$\lambda \neq 5$

Therefore, the system has a unique solution if $\lambda \neq 5$.

Note: The value of $\mu$ doesn't affect uniqueness.