Given,

$x=\alpha ,y=\beta ,z=\gamma$ be the unique solution of the system of simultaneous linear equations

$2x+3y-2z+4=0 .....\left(1\right)$

$3x-4y+3z+5=0 .....\left(2\right)$

$kx-2y+z+3=0 .......\left(3\right)$

Now solving the above equation with operation $\left(1\right)+2\times \left(3\right) \& \left(2\right)-3\times \left(3\right)$ we get,

$\left(2+2k\right)x-y=-10 ........\left(4\right)$

$\left(3-3k\right)x+2y=4 .....\left(5\right)$

Now eliminating $y$ by operation $2\times \left(4\right)+\left(5\right)$ we get,

$\left(4+4k+3-3k\right)x=-20+4$

$\Rightarrow x=-\frac{16}{k+7}$

Now given  $x=\alpha =-2$,

So $-2=-\frac{16}{k+7}$

$\Rightarrow 2k+14=16$

$\Rightarrow 2k=2$

Then $k=1$

Now solving options we get,

$\left|\begin{array}{cc}3& 5\\ 1& 2\end{array}\right|=3\times 2-5\times 1=1$