The set of values of K for which the system of equations $\begin{bmatrix}2 & 3 & 1\\4 & 5 & 0\\1 & K & 3\end{bmatrix} \begin{bmatrix}x\\ y \\z\end{bmatrix}=\begin{bmatrix}5 \\ 6 \\ 7 \end{bmatrix}$ gives a unique solution is :

$R-\begin{Bmatrix}\frac{11}{4}\end{Bmatrix}$

The correct answer is option (4) → $R-\begin{Bmatrix}\frac{11}{4}\end{Bmatrix}$

Unique sol. exists for $\begin{bmatrix}2 & 3 & 1\\4 & 5 & 0\\1 & K & 3\end{bmatrix}≠0$

$R_3→R_3-3R_1$

$\begin{bmatrix}2 & 3 & 1\\4 & 5 & 0\\-5 & K-9 & 0\end{bmatrix}≠0$

so $4K-36+25≠0⇒K≠\frac{11}{4}$

$K∉R-\{\frac{11}{4}\}$