The system of equations
$x + y + z = 4$
$x + 2y +3z= 12$
$x + 3y+λz = μ$ has a unique solution if

$λ≠5, μ$ can be any real number

The correct answer is Option (3) → $λ≠5, μ$ can be any real number

$\text{System: }x+y+z=4,\;x+2y+3z=12,\;x+3y+\lambda z=\mu$

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

$\det A=\begin{vmatrix}1&1&1\\1&2&3\\1&3&\lambda\end{vmatrix} =(2\lambda-9)-( \lambda-3)+1=\lambda-5$

$\text{Unique solution } \iff \det A\ne0 \iff \lambda\ne5$

$\lambda\ne5\;(\mu\in\mathbb{R})$