Find a basis of the transformations such that the $\mathcal{B}$-matrix is diagonal.
Orthogonal projection $T$ onto the plane $3x_1+x_2+2x_3=0$ in $\mathbb{R^3}$.
Reflection $T$ about the plane $x_1+2x_2+x_3=0$ in $\mathbb{R^3}$.
In both cases, the $\mathcal{B}$-matrix will be diagonal iff $T(v_i)=kv_i$ for all vectors $v_i$ in the basis.
For number 1, we know the plane is defined by $\left( \begin{array}{ccc}3 \\ 1\\2 \end{array} \right)$, and we choose our basis to be three vectors $v1 = \left( \begin{array}{ccc}3\\1\\2 \end{array} \right), v_2 = \left( \begin{array}{ccc}2\\0\\-3 \end{array} \right), v_3 = \left( \begin{array}{ccc}1\\-3\\0 \end{array} \right)$ so that $v_2$ and $v_3$ are perpendicular to $v_1$. Then we have $T(v_1) = v_1$, but I'm not sure how to calculate $T(v_2)$ and $T(v_3)$.
For number 2, we know the plane is defined by $\left( \begin{array}{ccc}1\\2\\1 \end{array} \right)$ and we again consider three vectors $v1 = \left( \begin{array}{ccc}1\\2\\1 \end{array} \right), v_2 = \left( \begin{array}{ccc}1\\0\\-1 \end{array} \right), v_3 = \left( \begin{array}{ccc}2\\-1\\0 \end{array} \right)$ so that $v_1$ is parallel to the plane and $v_2,v_3$ are perpendicular to the plane. In this case, I know $T(v_1)=v_1$, but again I'm not sure how to calculate $T(v_2)$ and $T(v_3)$.
In either case, is my initial work a correct procedure? Also, how do we calculate $T(v_2)$ and $T(v_3)$ in either case for $\mathbb{R^3}$. It doesn't seem so clear to see this geometrically as it is in $\mathbb{R^2}$.