Let $$a, b \in R$$ be such that the equation $$ax^2 - 2bx + 15 = 0$$ has repeated root $$\alpha$$ and if $$\alpha$$ and $$\beta$$ are the roots of the equation $$x^2 - 2bx + 21 = 0$$, then $$\alpha^2 + \beta^2$$ is equal to:

Given,

$$ax^2-2bx+15=0$$

has repeated root $$\alpha$$.

Hence, discriminant is zero:

$$(-2b)^2-4(a)(15)=0$$

$$4b^2-60a=0$$

$$b^2=15a$$

Now, for repeated root,

$$\alpha=\frac{2b}{2a}=\frac ba$$

Using

$$b^2=15a$$

we get

$$\alpha=\frac{15}{b}$$

Also, $$\alpha$$ and $$\beta$$ are roots of

$$x^2-2bx+21=0$$

Hence,

$$\alpha+\beta=2b,\qquad \alpha\beta=21$$

Since $$\alpha$$ is a root,

$$\alpha^2-2b\alpha+21=0$$

Substituting

$$\alpha=\frac{15}{b}$$

we get

$$\left(\frac{15}{b}\right)^2-2b\left(\frac{15}{b}\right)+21=0$$

$$\frac{225}{b^2}-30+21=0$$

$$\frac{225}{b^2}=9$$

$$b^2=25$$

Now,

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$

$$=(2b)^2-2(21)$$

$$=4(25)-42$$

$$=58$$

Hence, $$\boxed{58}$$.