Question

The constant $k$ is such that the following system of equations posses a non-trivial(i.e., not all zero) solution over the set of rationals $Q$
$x+ky+3z=0,3x+ky-2z=02x+3y-4z=0$
Then $\frac{2k}{11}$ is equal to

Answer 1

For non-trivial solution
$\Delta =\left|\begin{array}{ccc}1& k& 3\\ 3& k& -2\\ 2& 3& -4\end{array}\right|=0$
Expanding along first column,
$1(-4k+6)-3(-4k-9)+2(-2k-3k)=0\Rightarrow -2k+33=0\Rightarrow k=33/2$