Explanation

(d)
Given curve is $y = x$log$\ x$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = 1 + \log x$
The slope of the normal $= - \frac{1}{(dy/dx)} = \frac{- 1}{1 + \log x}$
The slope of the given line $2x - 2y = 3$ is1
Since, these lines are parallel
$\therefore\ \frac{- 1}{1 + \log x} = 1\ \ \Rightarrow \ \log x = - 2\ \ \Rightarrow \ \ \ x = e^{- 2}$
and $y = - 2e^{- 2}$
$\therefore$ Coordinates of the point are $(e^{- 2} - 2\ ,\ - 2e^{- 2})$