Can you help me find this unit vector? 
The vectors $a = (1,1,1)$ and $b = (−1, −1, −1)$ are given. Determine the unit vector $c$ such that $\angle (a, c) = \frac \pi 6$ and that the area of ​​the parallelogram constructed over the vectors $b$ and $c$ is equal to $\sqrt2.$

So I tried solving this like:
$$P=|b \times c|=\sqrt2 \Rightarrow b \times c=(y-z, z-x, x-y)$$
$$|c|=1 \Rightarrow x^2 +y^2+c^2=1$$
and $$∡ (a, c) = π / 6 \Rightarrow ac=|a||c|\cdot \cos ∡ (a, c)$$
And with this I got 3 equations but i really don't know if this is correct or not. Can you help me solve this, I need it for my homework.

As currently stated, the problem has no solution, indeed vectors $a$ and $b$ are aligned with lenght $\sqrt 3$ therefore for any vactor $c$ such that $\angle (a, c) = \frac \pi 6$ we have that the area of the parallelogram is
$$A=\sqrt 3 \cdot 1 \sin\left(\frac \pi 6\right)= \frac 3 2\neq \sqrt 2$$
