Consider the equality
\begin{align*} &\vec{a}\cdot\vec{b}+c=0 \\ \implies &\vec{a}\cdot(\vec{b}+\frac{\vec{a}}{\vec{a}\cdot\vec{a}}c)=0. \end{align*}
If the above equation is valid for any $\vec{a} $, can we say the following equation is valid?
\begin{align*} &\vec{b}+\frac{\vec{a}}{\vec{a}\cdot\vec{a}}c=0\\ \implies & \vec{b}=-\frac{c}{|\vec{a}\cdot\vec{a}|}\vec{a} \end{align*}
I feel something (in particular, with regard to sign) is wrong.
Can the above process be justified? If not, let me know the reason.
The answer is clearly wrong, because it says that $\vec b$ is proportional to $\vec a$, i.e. they must point in the same direction. But if you add to any such $\vec b_0$ a vector $\vec{b'}$ which is perpendicular to $\vec a$, then the product $\vec a . (\vec b_0+\vec{b'})$ is the same so your first equation still holds for any $\vec b=\vec b_0+\vec b'$. This equation tells you the component of $\vec b$ along $\vec a$, but the component of $\vec b$ perpendicular to $\vec a$ is completely arbitrary.
The error comes in the step (which you rightly have a question mark against) that your second equation is true for all $\vec a$, so the bracket must be zero. This works for $a f(b,a)=0\ \forall\ a \implies f(b,a)=0$ but not for $\vec a.\vec f=0 \ \forall \ \vec a \implies \vec f=0$. You cannot create 3 equations from 1 equation.
The glitch in @InertialObserver's proof is that their $\vec b$ is a different $\vec b$ for $i=1,2,3$.