If the system of linear equations x - 2y + kz = 1 2x + y + z = 2 3x...

$x - 2y + kz = 1$ .........(1)
$2x + y + z = 2$ .........(2)
$3x - y - kz = 3$ ........(3)

Hence,

\left[$\begin{cases}1 & - 2 & k \\ 2 & 1 & 1 \\ 3 & - 1 & - k\end{cases}$\right]\left[$$\begin{array}{ll}x \\ y \\ z\end{array}$$\right] = \left[$$\begin{array}{ll}1 \\ 2 \\ 3\end{array}$$\right]</p><p>For infinite solution,</p><p>\left|\begin{cases}1 & - 2 & k \\ 2 & 1 & 1 \\ 3 & - 1 & - k\end{cases}\right| = 0

$⇒ \left(- k + 1\right) + 2\left(- 2k - 3\right) + k\left(- 2 - 3\right) = 0$

$⇒ k = - \dfrac{1}{2}$

Put this value in the equation \left(1\right)$, we get</p><p>$2x - 4y - z = 2 . . . . \left(4\right)$</p><p>Adding the equation \left(4\right) \text{ and } \left(2\right)$, we get

$4x - 3y = 4$

Hence, the equation of straight line is $4x - 3y - 4 = 0$