Let the system of linear equations x + y + k z = 2 2 x + 3 y - z = 1 3 x + 4 y + 2 z = k have infinitely…

Linear equations $x+y+kz=2$, $2x+3y-z=1$ and $3x+4y+2z=k$ have infinitely many solutions.

Therefore,

$\left|\begin{array}{ccc}1& 1& k\\ 2& 3& -1\\ 3& 4& 2\end{array}\right|=0$

$\Rightarrow 1\left(10\right)-1\left(7\right)+k\left(-1\right)=0$

$\Rightarrow 10-7-k=0$

$\Rightarrow 3-k=0$

$\Rightarrow k=3$

For $k=3$, the second system of equations is

$4x+5y=7 \dots \left(1\right)$

$7x+8y=10 \dots \left(2\right)$

Clearly, they have a unique solution

Subtracting $\left(1\right) \& \left(2\right)$, we get

$3x+3y=3$

$\Rightarrow x+y=1$