Explanation

(d)
Let the line passing through the intersection of two lines is
$(x + y - 2) + \lambda(x - y) = 0$
or $(1 + \lambda)x + (1 - \lambda)y - 2 = 0$ …(i)
Which is parallel to $x + 2y = 5$
$\therefore - \frac{(1 + \lambda)}{(1 - \lambda)} = - \frac{1}{2}$
$\Rightarrow \ 2 + 2\lambda = 1 - \lambda\ \Rightarrow \ \lambda = - \frac{1}{3}$
On putting $\lambda = - \frac{1}{3}$ in Eq. (i), we get
$\frac{2}{3}x + \frac{4}{3}y - 2 = 0\ \Rightarrow \ x + 2y = 3$