Let  $$ x = \frac{\sqrt{13}  + \sqrt{11}}{\sqrt{13}  - \sqrt{11}}  and  y = \frac{1}{x} $$, then the value of  $$ 3x^2  - 5xy + 3y^2  is $$

$$ x = \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}  and  y = \frac{1}{x} $$

So, clearly from the above xy = 1

$$x+y =\frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}} + \frac{\sqrt{13} - \sqrt{11}}{\sqrt{13} + \sqrt{11}}$$

= $$ \frac{(\sqrt{13} + \sqrt{11})^2 + (\sqrt{13} - \sqrt{11})^2}{13-11}$$

= $$\frac {13+11+2\sqrt{143}+13+11-2\sqrt{143}} {2}$$

= $$\frac{48}{2}$$

= 24

So, x+y = 24

$$3x^2-5xy+3y^2 = 3(x+y)^2 - 11xy$$

=$$3(24)^2-11(1)$$

= $$ 3 \times 576 - 11$$

= 1728 - 11

= 1717