Given planes are:

$x-cy-bz=0$ ...(i)

$cx-y+az=0$ ...(ii)

$bx+ay-z=0$ ...(iii)

Equation of the plane passing through the line of intersection of the plane (i) and (ii) may be taken as:

$\left(x-cy-bz\right)+\lambda \left(cx-y+az\right)=0$

$or x\left(1+\lambda c\right)-y\left(c+\lambda \right)+z\left(-b+a\lambda \right)=0$ ...(iv)

If plane (iii) and (iv) are the same, then 

$\therefore \frac{1+c\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1} \mathrm{or}\mathrm{ }\lambda =-\frac{\left(a+bc\right)}{\left(ac+b\right)}$

and $\lambda =-\frac{(ab+c)}{\left(1-a^2\right)}$

$\therefore \frac{-\left(a+bc\right)}{\left(ac+b\right)}=-\frac{\left(ab+c\right)}{\left(1-a^2\right)}$

$\Rightarrow a-a^3+bc-a^{2}bc=a^{2}bc+a{c}^2+a{b}^2+bc$

$\Rightarrow 2a^{2}bc+a{b}^2+a{c}^2+a^3-a=0$

$\Rightarrow a\left(2abc+c^2+b^2+a^2-1\right)=0$

$\Rightarrow a^2+b^2+c^2+2abc=1$