Explanation

(c)
Given, $y = x\log_{e}x$
$\Rightarrow \ \ \ \ \frac{dy}{dx} = \frac{x}{x} + \log{x = 1 + \log x}$
$\therefore$ Slope of normal$= \frac{- 1}{\frac{dy}{dx}\ } = \frac{- 1}{1 + \log x}$
And given line is $2x - 2y + 3 = 0\ \Rightarrow \ \ \frac{dy}{dx} = 1$
Since, normal line is parallel to the given line
$\therefore\ \ \ - \frac{1}{1 + \log x} = 1\ \ \Rightarrow \ \ \ \ x = e^{- 2}$
Now, intersecting point of given curve and $x = e^{- 2}$ is $(e^{- 2},\ - 2e^{- 2})$
$\therefore$ Required equation of line is
$y + 2e^{- 2} = 1\left( x - e^{- 2} \right)\ \ \Rightarrow \ \ \ \ x - y = 3e^{- 2}$