D=∣314424534∣(R3→R3−2R1+3R2)
=∣310420530∣=0
Now, let P3≡4x+4y+4z−δ=0. If the system has solutions it will have infinite solution, so,P3≡αP1+βP2
Hence, 3α+β=4 and 4α+2β=4⇒α=2andβ=−2
So, for infinite solution 2μ−2=δ
For system to be inconsistent,
2μ=δ+2
So, checking option, (4,3) satisfies.