The value(s) of $K$, for which the system of linear equations $2x + y + z = 1,x+Ky-z=\frac{3}{2}$ and $3y-5z = 9$ does not possess a unique solution is

$K=\frac{7}{5}$

The correct answer is Option (1) → $K=\frac{7}{5}$

System: $\;2x+y+z=1,\;x+Ky-z=\frac{3}{2},\;3y-5z=9\,$.

Coefficient matrix $\;A=\begin{pmatrix}2&1&1\\[4pt]1&K&-1\\[4pt]0&3&-5\end{pmatrix}$.

$\det A=2\begin{vmatrix}K&-1\\[4pt]3&-5\end{vmatrix}-1\begin{vmatrix}1&-1\\[4pt]0&-5\end{vmatrix}+1\begin{vmatrix}1&K\\[4pt]0&3\end{vmatrix} =2(-5K+3)-(-5)+3=-10K+14\,$.

For no unique solution, $\det A=0\Rightarrow -10K+14=0\Rightarrow K=\frac{7}{5}$.

Answer: $K=\frac{7}{5}$