If the system of linear equations x + y + 3z = 0 x + 3y + k ^ 2 z =...

So D = 0 \to \left|\begin{cases}1 & 1 & 3 \\ 1 & 3 & {k}^{2} \\ 3 & 1 & 3\end{cases}\right| = 0 ⇒ {k}^{2} = 9

$x + y + 3z = 0$

$x + 3y + 9z = 0$

$3x + y + 3z = 0$

$( 1 ) - ( 3 )$
$x = 0 ⇒ y + 3z = 0$

$⇒ \dfrac{y}{z} = - 3$
$so, x + \left(\dfrac{y}{z}\right) = - 3$