Equation of tangent at (x₀, y₀) is T = 0

⇒ 3 xx₀ – 4yy₀ = –36

it should be parallel to 3x + 2y = 1

$\frac{3{\mathrm{x}}_0}{4{\mathrm{y}}_0}=\frac{-3}{2}$ ⇒ x₀ = – 2y₀

∵ (x₀ ,y₀) lies on curve ⇒ ₀3x² – ₀4y² = 36

⇒ 3(–2y₀)² – ₀4y² = 36 

⇒ ${\mathrm{y}}_0^2=\frac{9}{2}\Rightarrow {\mathrm{y}}_0=\pm \frac{3}{\sqrt{2}}\Rightarrow {\mathrm{x}}_0=\mp 3\sqrt{2}$

Point  ⇒ $\left(3\sqrt{2},-\frac{3}{\sqrt{2}}\right)$ or $\left(-3\sqrt{2},\frac{3}{\sqrt{2}}\right)$

Nearest point is $\left(3\sqrt{2},-\frac{3}{\sqrt{2}}\right)$

⇒ $\sqrt{2}\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)=\sqrt{2}\left(\left(-\frac{3}{\sqrt{2}}\right)-3\sqrt{2}\right)=-9$