Spans and Dot Product: Findin the linear combination Suppose $(v_1, v_2, v_3)$ is a set of vectors mutually perpendicular. Assume that $\|v_1\|= \sqrt{27}\quad \|v_2\| = \sqrt{14}\quad \|v_3\|= \sqrt{ 4}\ $
Let $w$ be a vector in $\operatorname{Span}(v_1, v_2, v_3)$ such that 
$w \cdot v_1 = -81\quad w \cdot v_2 = -28\quad w \cdot v_3 = -12\ $
Express $w$ as a linear combination of the vectors $(v_1, v_2, v_3)$.
Ok. So, basically, I equated the vector as follows:
$w=xv_1+y v_2+z v_3$
Then, I did the dot product on each side by w and $xv_1+y v_2+z v_3$.
I ended up with this:
$$\|w\|^2=x^2\|v_1\|^2+y^2 \|v_2\|^2+z^2 \|v_3\|^2$$
No idea where to go next. Please give me a hint if I'm on the right path. If yes, please give the next step.

Nah boy, You should look at it the other way. You have $$w=xv_1+y v_2+z v_3$$Now consider these $3$ equations, $$w.v_1=(xv_1+y v_2+z v_3).v_1=27x$$$$w.v_2=(xv_1+y v_2+z v_3).v_2=14y$$$$w.v_3=(xv_1+y v_2+z v_3).v_3=4z$$ That   gives you $x,y,z$.  Like $x=-3$, $y=-2, z=-3$
