Let $\mathrm{P}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$ be the point on the hyperbola $3{\mathrm{x}}^2-4{\mathrm{y}}^2=36$, which is nearest to the line $3\mathrm{x}+2\mathrm{y}=1$, Then $\sqrt{2}\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)$ is equal to 

Given hyperbola $3{\mathrm{x}}^2-4{\mathrm{y}}^2-36=0$
Diff. w.r.to x
$\begin{array}{l}6\mathrm{x}-8\mathrm{y}\cdot \frac{\mathrm{dy}}{\mathrm{dx}}=0\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3\mathrm{x}}{4\mathrm{y}}\end{array}$
Given ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{atp}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)}=\frac{-3}{2}$
$\begin{array}{r}\frac{3{\mathrm{x}}_0}{4{\mathrm{y}}_0}=\frac{-3}{2}\\ {\mathrm{x}}_0=-2{\mathrm{y}}_0\end{array}$
Since,$\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\text{ lie on }3{\mathrm{x}}^2-4{\mathrm{y}}^2-36=0$                  $\Rightarrow 3{\mathrm{x}}_0^2-4{\mathrm{y}}_0^2-36=0$
$\Rightarrow 3\left(4{\mathrm{y}}_0^2\right)-4{\mathrm{y}}_0^2=36\Rightarrow {\mathrm{y}}_0=\pm \frac{3}{\sqrt{2}}$
If ${\mathrm{y}}_0=\frac{3}{\sqrt{2}};{\mathrm{x}}_0=-3\sqrt{2}\therefore \text{ point }\mathrm{A}=\left(-3\sqrt{2},\frac{3}{\sqrt{2}}\right)$
Perpendicular distance from $\mathrm{A}\left(-3\sqrt{2},\frac{3}{\sqrt{2}}\right)$ to $3\mathrm{x}+2\mathrm{y}=1\text{ is }\frac{|-9\sqrt{2}+3\sqrt{2}-1|}{\sqrt{13}}=\frac{6\sqrt{2}+1}{\sqrt{13}}$
$\mathrm{If} {\mathrm{y}}_0=\frac{-3}{\sqrt{2}}\Rightarrow {\mathrm{x}}_0=3\sqrt{2}\text{ point }\mathrm{B}=\left(3\sqrt{2},\frac{-3}{\sqrt{2}}\right)$
Perpendicular distance from

 $$\begin{gathered} \mathrm{B}\left(3\sqrt{2},-\frac{3}{\sqrt{2}}\right)\text{ to }3\mathrm{x}+2\mathrm{y}-1=0\text{ is } \\ \frac{|9\sqrt{2}-3\sqrt{2}-1|}{\sqrt{13}}=\frac{|6\sqrt{2}-1|}{\sqrt{13}} \end{gathered}$$
Point B is the nearest to $3\mathrm{x}+2\mathrm{y}-1=0$
$\sqrt{2}\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)=\sqrt{2}\left({\mathrm{y}}_0+2{\mathrm{y}}_0\right)=3\sqrt{2}{\mathrm{y}}_0=-9$