Let ABC be an equilateral triangle and AD perpendicular to BC. Then AB$$^2$$ + BC$$^2$$ + CA$$^2$$ = ?

Given ABC is an equilateral triangle

Let the length of the side = a

$$=$$>  AB = BC = CA = a

AD is the perpendicular bisector to BC

$$=$$>  BD = CD = $$\frac{a}{2}$$

From $$\triangle$$ABD,

AD$$^2$$ + BD$$^2$$ = AB$$^2$$

$$=$$>  AD$$^2$$ + $$\left(\frac{a}{2}\right)^2$$ =  a$$^2$$

$$=$$>  AD$$^2$$ = a$$^2$$ - $$\frac{a^2}{4}$$

$$=$$>  AD$$^2$$ = $$\frac{3a^2}{4}$$

$$=$$>  $$a^2=\frac{4}{3}$$AD$$^2$$

$$\therefore\ $$AB$$^2$$ + BC$$^2$$ + CA$$^2$$ = a$$^2$$ + a$$^2$$ + a$$^2$$ = 3a$$^2$$ = $$3\left(\frac{3}{4}AD^2\right)$$ = $$4AD^2$$

Hence, the correct answer is Option C