Slope of a line passing through$P(2, 3)$ and intersecting the line, $x + y = 7$ at a distance of 4 units from$P$,

is

Let $x=2+r\cos \theta$ and 

$y=3+ r \sin \theta$ be points of

intersection.

Now, $x+y=7$                                 [Given] 

$\begin{array}{l}\Rightarrow 2+r\cos \theta +3+r\sin \theta =7\\ \Rightarrow r(\sin \theta +\cos \theta )=2\\ \Rightarrow \sin \theta +\cos \theta =\frac{2}{r}=\frac{2}{\pm 4}=\pm \frac{1}{2}\\ \Rightarrow (\sin \theta +\cos \theta {)}^2=\frac{1}{4}\end{array}$

$\Rightarrow 1+\sin 2\theta =\frac{1}{4}\Rightarrow \sin 2\theta =\frac{-3}{4}\Rightarrow \frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{-3}{4}$

$\Rightarrow \frac{2m}{1+m^2}=\frac{-3}{4}$, where$m$ is slope of required line 

$\begin{array}{r}\Rightarrow 3m^2+8m+3=0\\ \Rightarrow m=\frac{-8\pm \sqrt{64-36}}{6}=\frac{-4\pm \sqrt{7}}{3}\end{array}$

 Consider,$\frac{1-\sqrt{7}}{1+\sqrt{7}}=\frac{1-\sqrt{7}}{1+\sqrt{7}}\times \frac{1-\sqrt{7}}{1-\sqrt{7}}=\frac{(1-\sqrt{7}{)}^2}{1-7}$

$=\frac{1+7-2\sqrt{7}}{-6}=\frac{8\mid -2\sqrt{7}}{-6}=\frac{-4+\sqrt{7}}{3}$

Thus, option (d) satisfies the above condition.