Kristi, first of all if you are projecting $\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$ onto the $x_{1}=x_{2}$ plane, then you are projecting onto the plane $x=y$, not $y=z$ (since you defined $x=x_{1}$, $y=x_{2}$, and $z=x_{3}$).
Now, as you are trying to find the coordinates of the projection vector, imagine the geometric meaning -- $z$, the 'height' of the vector will not ever change, as it is not relevant to the equation, but $x$ and $y$ will, depending on where the vector lies. When we are trying to find a projection on an n-dimensional subspace $W$, we can use a formula of ${proj_{W}}{\vec{x}}$=$(\vec{u_1}\cdot \vec{x})$\vec{u_1}$+$(\vec{u_2}\cdot \vec{x})$\vec{u_2}$+\cdots +$(\vec{u_n}\cdot \vec{x})$\vec{u_n}$, where $\vec{u_1}, \vec{u_2}\dots \vec{u_n}$ form an orthonormal basis of the subspace $W$. Here, $W$ is defined as $x=y$, meaning it can be spanned by vectors $\vec{v_1}$ = $ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ and $\vec{v_2}$ = $ \begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$, for example. To find an orthonormal basis of our space (meaning that all vectors in it will be mutually orthogonal/perpendicular, as well as of a length one), let's use the Gram-Schmidt process. An orthonormalized version of the vector $ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ would be $\vec{u_1}$ = $\frac{1}{\sqrt{2}}$ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$, as that will make it of length one. Now, by Gram-Schmidt, $\vec{u_2}=\vec{v_2}-\frac{\vec{v_2}\cdot\vec{u_1}}{\vec{u_1}\cdot\vec{u_1}} \vec{u_1}$, since we are basically subtracting the $\vec{u_1}$ component from our second vector, in order to get a vector perpendicular to $\vec{u_1}$ as a result. Calculations result into the following: $\vec{u_2}$= $\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$ $-\frac{\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}}{\frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}} \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ = $\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$ - $\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$=$\begin{bmatrix} 0\\ 0\\ 2 \end{bmatrix}$. Normalizing the resulting vector, we get $\vec{u_2}$ = $\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$.
Now that we have an orthogonal basis $\vec{u_1}$ = $\frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ and $\vec{u_2}$ = $\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$, we can calculate the projection.
So, to find the projection of your vector $\begin{bmatrix} x_1\\\ x_2\\\ x_3 \end{bmatrix}$ we use our orthonormal basis and the projection formula: ${proj_{W}}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$=($\frac{1}{\sqrt{2}}$\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} \cdot \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix})(\frac{1}{\sqrt{2}}$\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix})$+$(\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \cdot \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix})(\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix})$. After arithmetic, this results into ${proj_{W}}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$=$(\frac{x_1+x_2}{\sqrt{2}})$$(\frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix})$+$\begin{bmatrix} 0\\ 0\\ x_3 \end{bmatrix}$=$\begin{bmatrix} \frac{x_1+x_2}{2}\\ \frac{x_1+x_2}{2}\\ x_3 \end{bmatrix}$.
So, now you have your coordinates.
To find out the eigenvalues, think of the nature of the transformation -- the projection will not do anything to a vector if it is within the plane onto which you are projecting, and it will crash it if the vector is perpendicular to the plane. So, your eigenvalues are 1 and 0. A basis of eigenspace of 1 $\xi_{1}$ will have two vectors, as the plane is spanned by two of them. You could choose them to be your original $v_1$ and $v_2$, which were $ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$. To find a basis of eigenspace of 0 $\xi_{0}$, you need to find a vector perpendicular to this plane. You could use a property of cross-product, which states that $\vec{v_1} \times \vec{v_2}$ produces a vector $\vec{v_3}$ perpendicular to both. Crossing the aforementioned vectors, you get $\vec{v_3}=\begin{bmatrix} 2\\ -2\\ 0 \end{bmatrix}$.
Now that you know all of this, finding the matrix B is very easy by inspection; consider $\begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}$.