Write the transformation as a map $ \ T:\mathbb{R}^3 \to \mathbb{R}^3 \ $ Let $ \ T: \mathbb{R}^3 \to \mathbb{R}^3 \ $ be the linear transformation obtained by $ \ 120^{\circ} \ $ rotation about the line $ l: \ x=\lambda , \ y=\lambda , \ z=-\lambda \ ; \ \ where \ \  \lambda \in \mathbb{R} $ 
(a) Write the  transformation as a map $ \ T:\mathbb{R}^3 \to \mathbb{R}^3 \ $ given by $ \ T(\vec u)=A \cdot \vec u \ $
(b) Find the image of the plane $ \ P: \ x+2y-2z=6 \ $ under the transformation $ \ T \ $. What Geometrical object do you obtain?
Answer:
I need the linear transformation $ \ T \ $ obtained by $ \ 120^{\circ} \ $ rotation about the line $ \ l: x=\lambda , \ y=\lambda , \ z=-\lambda \ $
I know the rotation matrix about a point , about an axis but  not about a line
I need help right here.
 A: We first set up a new orthonormal basis $(f_1,f_2,f_3)$ adapted to the given $T$. To this end choose $$f_3:={1\over\sqrt{3}}(1,1,-1), \quad f_1:={1\over\sqrt{2}}(-1,1,0), \qquad f_2:=f_3\times f_1={1\over\sqrt{6}}(1,1,2)\ .$$
The transform matrix from the standard basis $(e_1,e_2,e_3)$ to $(f_1,f_2,f_3)$ then is
$$S:=\left[\matrix{-{1\over\sqrt{2}}&{1\over\sqrt{6}}&{1\over\sqrt{3}}\cr
{1\over\sqrt{2}}&{1\over\sqrt{6}}&{1\over\sqrt{3}}\cr
0&{2\over\sqrt{6}}&-{1\over\sqrt{3}}\cr}\right]\ .$$
With respect to the basis $f:=(f_1,f_2,f_3)$ the given transformation $T$ has the matrix
$$[T]_f=\left[\matrix{-{1\over2}&-{\sqrt{3}\over2}&0\cr
{\sqrt{3}\over2}&-{1\over2}&0\cr
0&0&1\cr}\right]$$
(we have chosen the sense of rotation here). According to the rules of linear algebra the matrix of $T$ with respect to the standard basis then is given by
$$[T]_e=S\> [T]_f\>S^\top\ .$$
As for the second problem: Coordinatewise the map ${\bf x}\mapsto {\bf x}':=T{\bf x}$ is given by
$${\bf x'}=[T]_e\>{\bf x}\ .$$
This implies
$${\bf x}=[T]_e^{-1}{\bf x}'=[T]_e^\top{\bf x}'\ .\tag{1}$$
This allows to express the coordinates $(x,y,z)$ of a preimage point ${\bf x}$ by the coordinates $(x',y',z')$ of the image point ${\bf x}'$. The point ${\bf x}$ lies in the plane $P$ iff its coordinates satisfy $x+2y-2z=6$. Plug the  expressions in terms of $x'$, $y'$, $z'$ resulting from $(1)$ into this equation, and you obtain the equation of the image plane $T(P)$.
A: $R=\begin{pmatrix}\cos\frac{2\pi}3&\sin\frac{2\pi}3&0\\-\sin\frac{2\pi}3&\cos\frac{2\pi}3&0\\0&0&1\end{pmatrix}$ is rotation about the $z$-axis.  Similarly one can do the $x$ and $y$ axes...
We will form an orthogonal matrix,$$R_1=\begin{pmatrix}\frac1{\sqrt6}&\frac1{\sqrt6} &\frac2{\sqrt6}\\\frac1{\sqrt2}&-\frac1{\sqrt2}&0\\\frac1{\sqrt3}&\frac1{\sqrt3}&-\frac1{\sqrt3}\end{pmatrix}$$.
The top row will be the cross-product of the other two.   This matrix makes the last row the new $z$-axis. (Note:  there's a question of orientation here.   I chose $(1,1,-1)$.  The opposite vector could have been chosen.)
Finally,  the rotation will be given by $R_1^{-1}RR_1$... 
Of course,  $(R_1)^{-1}=R_1^t$.
Check, for instance,  that $(2,2,-2)$ is fixed...
For part $2$, the image should be another plane (the rotation of a plane is a plane).
