$$x + y + kz = 2$$ ............(i)

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

$$3x + 4y + 2z = k$$ ......(iii)

(1) + (2)

$$3x + 4y + z(k - 1) = 3$$

Comparing with (3)

$$k = 3$$

Now, $$4x + 5y = 7$$

$$ \Rightarrow 3x + 3y = 3$$

$$7x + 8y = 10$$

as $${4 \over 7} \ne {5 \over 8}$$

$$\therefore$$ Unique solution satisfying $$x + y = 1$$