1. Let the perpendicular from the origin to the line be the normal to the line.

   Any line at perpendicular distance $p$ from the origin can be written in normal form as $x\cos\alpha + y\sin\alpha = p,$ where $\alpha$ is the angle made by the perpendicular (normal) from the origin with the positive $x$-axis.

   Here, the distance from the origin is $4$, so the line is $x\cos\alpha + y\sin\alpha = 4.$

2. Use the angle condition.

   The line $x+y=0$ has slope $-1$, so it makes an angle $-45^\circ$ (or $135^\circ$) with the positive $x$-axis.

   The perpendicular from the origin to $L$ makes an angle $60^\circ$ with the line $x+y=0$.

   Hence possible directions for the normal are: $-45^\circ + 60^\circ = 15^\circ,$ or $-45^\circ - 60^\circ = -105^\circ,$ equivalently $255^\circ$.

   Since the line makes positive intercepts on both axes, in the equation $x\cos\alpha + y\sin\alpha = 4,$ we need both $\cos\alpha > 0$ and $\sin\alpha > 0$.

   So $\alpha$ must lie in the first quadrant. Therefore, $\alpha = 15^\circ.$

3. Substitute $\alpha=15^\circ$.

   Thus the line is $x\cos 15^\circ + y\sin 15^\circ = 4.$

   Now, $\cos 15^\circ = \frac{\sqrt{3}+1}{2\sqrt{2}}, \qquad \sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}}.$

   Hence, $x\cdot \frac{\sqrt{3}+1}{2\sqrt{2}} + y\cdot \frac{\sqrt{3}-1}{2\sqrt{2}} = 4.$

4. Simplify.

   Multiply throughout by $2\sqrt{2}$: $(\sqrt{3}+1)x + (\sqrt{3}-1)y = 8\sqrt{2}.$

5. Match with the options.

   This is exactly Option C.

6. Check positive intercepts.

   • $x$-intercept: set $y=0$ $x = \frac{8\sqrt{2}}{\sqrt{3}+1} > 0$
   • $y$-intercept: set $x=0$ $y = \frac{8\sqrt{2}}{\sqrt{3}-1} > 0$

   Both are positive, so the condition is satisfied.

Therefore, the correct equation is $(\sqrt{3}+1)x + (\sqrt{3}-1)y = 8\sqrt{2}.$