Vectors : finding scalar $\mu$ in $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $ 
Non-zero vectors $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ satisfy $\overrightarrow a .\overrightarrow b  = 0$, $\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow b  + \overrightarrow c } \right) = 0$ and $2\left| {\overrightarrow b  + \overrightarrow c } \right| = \left| {\overrightarrow b  - \overrightarrow a } \right|$. If $\overrightarrow a  = \mu \overrightarrow b  + 4\overrightarrow c $, then find all possible values of $\mu $.

My approach is as follows
Given $\left( {\overrightarrow b  - \overrightarrow a } \right).\left( {\overrightarrow b  + \overrightarrow c } \right) = 0\& 2\left| {\overrightarrow b  + \overrightarrow c } \right| = \left| {\overrightarrow b  - \overrightarrow a } \right|$
$\overrightarrow u  = \left( {\overrightarrow b  - \overrightarrow a } \right);\overrightarrow v  = \left( {\overrightarrow b  + \overrightarrow c } \right)$
$2\left| {\overrightarrow v } \right| = \left| {\overrightarrow u } \right| \Rightarrow 4\left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow c } \right|}^2} + 2\overrightarrow b .\overrightarrow c } \right) = \left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow a } \right|}^2} - 2\overrightarrow a .\overrightarrow c } \right)$
$\overrightarrow a  = \mu \overrightarrow b  + 4\overrightarrow c  \Rightarrow \overrightarrow b  - \overrightarrow u  = \mu \overrightarrow b  + 4\left( {\overrightarrow v  - \overrightarrow b } \right)$
$ \Rightarrow \overrightarrow b .\overrightarrow v  - \overrightarrow u .\overrightarrow v  = \mu \overrightarrow b .\overrightarrow v  + 4\left( {\overrightarrow v .\overrightarrow v  - \overrightarrow b .\overrightarrow v } \right) \Rightarrow 5\overrightarrow b .\overrightarrow v  = \mu \overrightarrow b .\overrightarrow v  + 4{\left| {\overrightarrow v } \right|^2}$\
$ \Rightarrow 5\overrightarrow b .\left( {\overrightarrow b  + \overrightarrow c } \right) = \mu \overrightarrow b .\left( {\overrightarrow b  + \overrightarrow c } \right) + 4{\left| {\overrightarrow v } \right|^2} \Rightarrow 5\overrightarrow b .\overrightarrow b  + 5\overrightarrow b .\overrightarrow c  = \mu \overrightarrow b .\overrightarrow b  + \mu \overrightarrow b .\overrightarrow c  + 4{\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 8\overrightarrow b .\overrightarrow c $
Not able to proceed from here onward

As $\vec a \perp \vec b$ like $\hat i \perp \hat j$, it's strategic to represent
$$ \vec c = \frac{1}{4}\vec a - \frac{\mu}{4}\vec b$$
Remaining two conditions are,
$$ (\vec b - \vec a)\cdot(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)=0$$
and
$$ (\vec b - \vec a)\cdot(\vec b - \vec a) = 2^2(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)\cdot(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)$$
which should give you a quadratic in $\mu$.
