The value of k for which the system of equations
$x+y+z=1$
$x-ky+z=1$
$x-y+z=1$
has more than one solutions is

any real number

The correct answer is Option (4) → any real number

Given equations:

$x + y + z = 1$

$x - ky + z = 1$

$x - y + z = 1$

Subtract $(3^{rd})$ from $(1^{st})$:

$(x + y + z) - (x - y + z) = 1 - 1$

$2y = 0 \Rightarrow y = 0$

Substitute $y = 0$ in $(1^{st})$:

$x + z = 1$ ...(i)

Substitute $y = 0$ in $(2^{nd})$:

$x + z = 1$ ...(ii)

Equations (i) and (ii) are identical.

Hence, the system has infinitely many solutions for all $k \in \mathbb{R}$.

Therefore, $k$ can be any real number.