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 seek the point $$P(x_0, y_0)$$ on the hyperbola $$3x^2 - 4y^2 = 36$$ that is closest to the line $$3x + 2y = 1$$, and then compute $$\sqrt{2}(y_0 - x_0)\,.$$

First, rewrite the hyperbola in standard form as $$\frac{x^2}{12} - \frac{y^2}{9} = 1\,, $$ so that $$a^2 = 12$$ and $$b^2 = 9\,. $$ A convenient parametrization is for the right branch $$x = 2\sqrt{3}\cosh t,\quad y = 3\sinh t$$ and for the left branch $$x = -2\sqrt{3}\cosh t,\quad y = 3\sinh t\,. $$

The distance from a point $$(x,y)$$ to the line $$3x + 2y - 1 = 0$$ is given by $$d = \frac{\bigl|3x + 2y - 1\bigr|}{\sqrt{13}}\,. $$ On the right branch we set $$f(t) = 6\sqrt{3}\cosh t + 6\sinh t - 1$$ since $$3x + 2y - 1 = 6\sqrt{3}\cosh t + 6\sinh t - 1\,. $$ Differentiating yields $$f'(t) = 6\sqrt{3}\sinh t + 6\cosh t = 0 \quad\Longrightarrow\quad \tanh t = -\frac{1}{\sqrt{3}}\,. $$ Together with $$\cosh^2 t - \sinh^2 t = 1$$ one finds $$\cosh t = \sqrt{\frac{3}{2}},\qquad \sinh t = -\frac{1}{\sqrt{2}}\,. $$ Hence the critical point on the right branch is $$x_0 = 2\sqrt{3}\,\sqrt{\tfrac{3}{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2},\qquad y_0 = 3\Bigl(-\frac{1}{\sqrt{2}}\Bigr) = -\frac{3}{\sqrt{2}} = -\frac{3\sqrt{2}}{2}\,. $$

On the left branch we instead have $$f(t) = -6\sqrt{3}\cosh t + 6\sinh t - 1\,, $$ and setting $$f'(t)=0$$ gives $$\tanh t = \frac{1}{\sqrt{3}}\,, $$ leading to $$x_0 = -3\sqrt{2},\qquad y_0 = \frac{3\sqrt{2}}{2}\,. $$

To decide which point is nearer, compute the numerators of the distances. For the right‐branch point $$\bigl(3\sqrt{2},-\tfrac{3\sqrt{2}}{2}\bigr)$$ one has $$\bigl|3(3\sqrt{2}) + 2\bigl(-\tfrac{3\sqrt{2}}{2}\bigr) - 1\bigr| = \bigl|9\sqrt{2} - 3\sqrt{2} - 1\bigr| = \bigl|6\sqrt{2} - 1\bigr|\approx 7.49\,. $$ For the left‐branch point $$\bigl(-3\sqrt{2},\tfrac{3\sqrt{2}}{2}\bigr)$$ one gets $$\bigl|3(-3\sqrt{2}) + 2\bigl(\tfrac{3\sqrt{2}}{2}\bigr) - 1\bigr| = \bigl|-9\sqrt{2} + 3\sqrt{2} - 1\bigr| = \bigl|-6\sqrt{2} - 1\bigr|\approx 9.49\,. $$ Thus the right branch point is closer to the line.

Finally, at that nearest point we compute $$\sqrt{2}(y_0 - x_0) = \sqrt{2}\Bigl(-\frac{3\sqrt{2}}{2} - 3\sqrt{2}\Bigr) = \sqrt{2}\,\Bigl(-\tfrac{9\sqrt{2}}{2}\Bigr) = -\frac{9\times 2}{2} = -9\,. $$ The correct answer is Option C: $$-9\,. $$