If $$x^{2a} = y^{2b} = z^{2c}$$ ≠ 0 and $$x^2 = yz$$, then the value of $$\frac{ab + bc + ca}{bc}$$ is:

$$x^{2a} = y^{2b} = z^{2c}$$ = k

$$x^{2a} = k$$

$$x = (k)^{\frac{1}{2a}}$$

Likewise,

$$y = (k)^{\frac{1}{2b}}$$

$$z = (k)^{\frac{1}{2c}}$$

Now,

$$x^2 = yz$$

$$((k)^{\frac{1}{2a}})^2 = (k)^{\frac{1}{2b}}(k)^{\frac{1}{2c}}$$

$$(k)^{\frac{1}{a}} = (k)^{\frac{1}{2b} + \frac{1}{2c}}$$

$$\frac{1}{a} = \frac{1}{2b} + \frac{1}{2c}$$

By adding 1/2a both sides,

$$\frac{1}{a} + \frac{1}{2a} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}$$

$$\frac{3}{2a} = \frac{ab + bc + ac}{2abc}$$

$$\frac{ab + bc + ac}{bc} = 3$$