Mathematics - Determinant Question with Solution | TestHub

We have,

$3x-y+4z=3$

$x+2y-3z=-2$

$6x+5y+kz=-3$

For infinitely many solutions,

$\left|A\right|=0$

$\Rightarrow \left|\begin{array}{ccc}3& -1& 4\\ 1& 2& -3\\ 6& 5& k\end{array}\right|=0$

$\Rightarrow 3(2k+15)+1\left(k+18\right)+4\left(5-12\right)=0$

$\Rightarrow 3(2k+15)+k+18-28=0$

$\Rightarrow 7k+35=0$

$\Rightarrow k=-5$