The distance of the point ( 2 , 3 ) from the line 2 x - 3 y + 28 = 0 , measured parallel to the line 3 x - y…

Let, $L_1:\sqrt{3}x-y+1=0$ and $L_2:2x-3y+28=0$.

Slope, $m_1=\sqrt{3}=\tan \theta$

$\Rightarrow \theta ={60}^{°}$

$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}, \cos \theta =\frac{1}{2}$

Now, equation of line passing through $\left(2,3\right)$ and having slope $\sqrt{3}$ is,

$\frac{x-2}{{\displaystyle \frac{1}{2}}}=\frac{{\displaystyle y-3}}{{\displaystyle \frac{\sqrt{3}}{2}}}=r$

$\Rightarrow 2x-3y+28=0$

$\Rightarrow 4+r-9-\frac{3\sqrt{3}}{2}r+28=0$

$\Rightarrow \left(\frac{2-3\sqrt{3}}{2}\right)r+23=0$

$\Rightarrow r=\frac{46}{3\sqrt{3}-2}\times \frac{3\sqrt{3}+2}{3\sqrt{3}+2}$

$\Rightarrow r=\frac{46}{23}\times \frac{3\sqrt{3}+2}{1}=4+6\sqrt{3}$