<s>[INST] 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 $a,b,c$?
 [/INST] The Rodrigues' Rotation Matrix Formula states that
$R = {a a}^T + (I - {a a}^T) \cos \theta + S_a \sin \theta $
where $a$ is the unit vector of the axis, and
$ S_a = \begin{bmatrix} 0 && - a_z && a_y \\ a_z && 0 && -a_x \\ -a_y && a_x && 0 \end{bmatrix} $
Multiply this matrix by $e_2 = [0, 1, 0]^T $ to get
$[a, b, c]^T = {a a}^T e_2 + (I - {a a}^T) e_2 \cos \theta + S_a e_2 \sin \theta$
The rest is just computation, we have
$ {a a}^T = \dfrac{1}{3} \begin{bmatrix} 1 && 1 && 1 \\ 1 && 1 && 1 \\ 1 && 1 && 1 \end{bmatrix}$
$ I - {a a}^T = \dfrac{1}{3} \begin{bmatrix} 2 && -1 && -1 \\ -1 && 2 && -1 \\ -1 && -1 && 2 \end{bmatrix} $
$S_a = \dfrac{1}{\sqrt{3}} \begin{bmatrix} 0 && -1 && 1 \\ 1 && 0 && -1 \\ -1 && 1 && 0 \end{bmatrix}    $
Therefore,
$ \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \dfrac{1}{6} \begin{bmatrix} 1 \\ 4 \\ 1 \end{bmatrix} + \dfrac{1}{2} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = \dfrac{1}{6} \begin{bmatrix} -2 \\ 4 \\ 4 \end{bmatrix} $
Therefore, $ 3 a^2 + b^2 + c^2 = \dfrac{1}{36} (12 + 16 + 16) = \dfrac{44}{36} = \dfrac{11}{9} $
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