Rotation matrix from given axis Let $T\colon\mathbb{R^3}\to\mathbb{R^3}$ denotes linear transformation which rotates by $\frac{\pi}{3}$ counter-clockwise along the vector $u=(1,1,1)$.
If $T(0,1,0)=(a,b,c)$, Find $3a^2+b^2+c^2$.

My Attempt
Consider a plane which contains a point $(0,1,0)$ and uses $u$ as normal vector.
Then I got : $x+y+z=1$.
Since $T$ is rotation, $T(0,1,0)=(a,b,c)$ must be on the same plane $x+y+z=1$.
This means $a+b+c=1$.
Again, $T$ preserves norm of vector : $1=|(0,1,0)|=|T(a,b,c)|=a^2+b^2+c^2$.
So, I got two equations :
$$a+b+c=1 \\ a^2+b^2+c^2=1$$
I want to Find $a, b, c$ without finding exact form of $T$ but I need one more equation about $a,b,c$ to solve this system.
Is there any property of rotation which can give me one more equation about $a,b,c$?

Another ad-hoc approach. Passing to the plane $x+y+z=1$ where the action takes place, we can see that $(0,1,0)$ rotates around $\left(\frac13, \frac13, \frac13\right)$ within the plane by the angle $\frac{\pi}3$ counter-clockwise.
By definition of $T$ we have

*

*The angle $\angle (0,1,0),\left(\frac13, \frac13, \frac13\right),(a,b,c)$ is equal to $\frac{\pi}3$,

*The distances from $\left(\frac13, \frac13, \frac13\right)$ to the points $(a,b,c)$ and $(0,1,0)$ are equal.

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It follows that the triangle with vertices $(0,1,0),\left(\frac13, \frac13, \frac13\right),(a,b,c)$ is equilateral so in particular
\begin{align}
\left(a-\frac13\right)^2 + \left(b-\frac13\right)^2 + \left(c-\frac13\right)^2 &= d\left(\left(\frac13, \frac13, \frac13\right), (a,b,c)\right)^2 \\
&= d\left((0,1,0),(a,b,c)\right) \\
&= a^2+(b-1)^2+c^2
\end{align}
which is equivalent to $a-2b+c=1$, so this is your third equation as stated by @enzotib.
You can further simplify the system of equations if you show that the triangle with vertices $(0,0,1),\left(\frac13, \frac13, \frac13\right),(a,b,c)$ is also equilateral (and hence congruent to the aforementioned one) so in particular we have
$$(a-1)^2+b^2 + c^2 = d((0,0,1),(a,b,c)) = d((0,1,0),(a,b,c)) = a^2+(b-1)^2+c^2$$
which immediately implies $a=b$, eliminating one of the solutions.
