Let $$P(x_0, y_0)$$ be the point on the hyperbola $$3x^2 - 4y^2 = 36$$, which is nearest to the line $$3x + 2y = 1$$. Then $$\sqrt{2}(y_0 - x_0)$$ is equal to :

We need to find the point $$P(x_0, y_0)$$ on the hyperbola $$3x^2 - 4y^2 = 36$$ nearest to the line $$3x + 2y = 1$$. The nearest point occurs where the tangent to the hyperbola is parallel to this line, whose slope is $$-\frac{3}{2}$$. Differentiating the hyperbola gives $$6x - 8y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{3x}{4y}$$, and setting this equal to $$-\frac{3}{2}$$ yields $$\frac{3x_0}{4y_0} = -\frac{3}{2} \implies x_0 = -2y_0$$.

Substituting into $$3x^2 - 4y^2 = 36$$ gives $$3(-2y_0)^2 - 4y_0^2 = 36$$, i.e.\ $$12y_0^2 - 4y_0^2 = 36 \implies 8y_0^2 = 36 \implies y_0^2 = \frac{9}{2}$$. Hence $$y_0 = \pm \frac{3}{\sqrt{2}}$$ and correspondingly $$x_0 = \mp \frac{6}{\sqrt{2}} = \mp 3\sqrt{2}$$.

To determine which of the points $$(3\sqrt{2}, -\frac{3}{\sqrt{2}})$$ and $$(-3\sqrt{2}, \frac{3}{\sqrt{2}})$$ is nearer to the line $$3x + 2y - 1 = 0$$, we compute for $$(3\sqrt{2}, -\frac{3}{\sqrt{2}})$$ the distance $$\frac{|9\sqrt{2} - \frac{6}{\sqrt{2}} - 1|}{\sqrt{13}} = \frac{|9\sqrt{2} - 3\sqrt{2} - 1|}{\sqrt{13}} = \frac{|6\sqrt{2} - 1|}{\sqrt{13}}$$ and for $$(-3\sqrt{2}, \frac{3}{\sqrt{2}})$$ the distance $$\frac{|-9\sqrt{2} + 3\sqrt{2} - 1|}{\sqrt{13}} = \frac{|6\sqrt{2} + 1|}{\sqrt{13}}$$, so the first point is nearer.

Finally, computing $$\sqrt{2}(y_0 - x_0)$$ gives $$\sqrt{2}\left(-\frac{3}{\sqrt{2}} - 3\sqrt{2}\right) = \sqrt{2} \cdot \left(-\frac{3}{\sqrt{2}} - 3\sqrt{2}\right) = -3 - 6 = -9$$.

Answer: -9