If the system of equations 2x + 3y - z = 0, x + ky - 2z = 0 and 2x...

The system of equations ${a}_{1}x + {b}_{1}y + {c}_{1}z = 0, {a}_{2}x + {b}_{2}y + {c}_{2}z = 0 and${a}_{3}x + {b}_{3}y + {c}_{3}z = 0has a non-trivial solution, if </p><p>\left|\begin{cases}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{cases}\right| = 0 .

Hence, for the given lines $2x + 3y - z = 0, x + ky - 2z = 0$ and $2x - y + z = 0,$ we have

\left|$\begin{cases}2 & 3 & - 1 \\ 1 & k & - 2 \\ 2 & - 1 & 1\end{cases}$\right| = 0$</p><p>$⇒ 2\left(k - 2\right) - 3\left(1 + 4\right) - 1\left(- 1 - 2k\right) = 0$</p><p>$⇒ 2k - 4 - 15 + 1 + 2k = 0$</p><p>$⇒ k = \dfrac{9}{2}$</p><p>Thus, the lines are </p><p>$2x + 3y - z = 0 . . . \left(1\right)$</p><p>$x + \dfrac{9}{2}y - 2z = 0 . . . \left(2\right)$</p><p>$2x - y + z = 0 . . . \left(3\right)$</p><p>\left(1\right) - \left(3\right) ⇒ 4y - 2z = 0$

$2y = z . . . \left(4\right)$

$\dfrac{y}{z} = \dfrac{1}{2} . . . \left(5\right)$

put $z$ from eq \left(4\right)$into \left(1\right)$

$2x + 3y - 2y = 0$

$2x + y = 0$

$\dfrac{x}{y} = - \dfrac{1}{2} . . . \left(6\right)$

From eq \left(4\right)$and \left(6\right)$

$\dfrac{z}{x} = - 4$

Hence, $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} + k = - \dfrac{1}{2} + \dfrac{1}{2} - 4 + \dfrac{9}{2} = \dfrac{1}{2} .$