# Solving vector equation employing products.

If $$\vec{a}\times(\vec{b}\times\vec{c})+(\vec{a}\cdot\vec{b})\vec{b}=(4-2\lambda-\sin\alpha)\vec{b}+(\lambda^2-1)\vec{c}$$ and $(\vec{c}\cdot\vec{c})\vec{a}=\vec{c}$ where $\vec{b},\vec{c}$ are collinear. Then find values of $\lambda$ and $\alpha$.

My tries:

$(\vec{c}\cdot\vec{c})\vec{a}=\vec{c}\implies (\vec{c}\cdot\vec{c})\vec{a}\times\vec{c}=\vec{c}\times\vec{c}=0\implies \vec{a}\times \vec{c}=0\implies \vec{a}\parallel\vec{c}\parallel\vec{b}\tag{1}$

Also $(\vec{c}\cdot\vec{c})\vec{a}=\vec{c}\implies |\vec{c}|^2|\vec{a}|=|\vec{c}|\implies |\vec{c}||\vec{a}|=1$ this along with $(1)$

$\implies \vec{a}\cdot\vec{c}=1 \tag{2}$

From $(1)$, $\vec{a}\times(\vec{b}\times\vec{c})=0$ also $\vec{b}=\mu\vec{c}$ for some real $\mu$.

$\implies (\vec{a}\cdot\vec{b})\vec{b}=(4-2\lambda-\sin\alpha)\vec{b}+(\lambda^2-1)\vec{c}\\ \implies(\vec{a}\cdot\vec{\mu\vec{c}})\vec{\mu\vec{c}}=(4-2\lambda-\sin\alpha)\vec{\mu\vec{c}}+(\lambda^2-1)\vec{c}$

Taking dot with $\vec{a}$ then from $(2)$ $\implies \mu^2-(4-2\lambda-\sin\alpha)\mu+1-\lambda^2=0$

Now only one such unique real $\mu$ should exist, making discriminant $0$ won't helped me.

What should I do? Also, I even don't know whether above is correct or not as I'm very new to this course. Please help.