For which of the following ordered pairs μ , δ , the system of linear equations x + 2 y + 3 z = 1…

$D=\left|\begin{array}{ccc}3& 4& 5\\ 1& 2& 3\\ 4& 4& 4\end{array}\hspace{0.17em}\right| \left(R_3\to R_3-2R_1+3R_2\right)$
    $=\left|\begin{array}{ccc}3& 4& 5\\ 1& 2& 3\\ 0& 0& 0\end{array}\hspace{0.17em}\right|=0$
Now, let $P_3\equiv 4x+4y+4z-\delta =0.$ If the system has solutions it will have infinite solution, so,$P_3\equiv \alpha P_1+\beta P_2$
Hence, $3\alpha +\beta =4$ and $4\alpha +2\beta =4\Rightarrow \alpha =2\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\beta =-2$
So, for infinite solution $2\mu -2=\delta$

For system to be inconsistent,

$\hspace{0.17em}2\mu \ne \delta +2$
So, checking option, $(4, 3)$ satisfies.