A half-projection into a plane is a linear transformation that fixes the plane and moves all other points half way to the plane (instead of all the way, in a standard projection.)
Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the (orthogonal) half-projection into the following plane:
$$3x - 3y + 3z = 0$$
Find the matrix of this linear transformation.
That is all that is provided.
On
From the given equation $(3,-3,3)$ is perpendicular to the plane, hence you want to map it to $(\frac32,-\frac32,\frac32)$ By educated guessing, $(1,1,0)$ and $(0,1,1)$ are in the plane and hence you want to map these to themselves. Can you find the matrix from these three conditions?
On
Note that the plane $3x-3y+3z=0$ consists of those vectors in $\mathbb{R}^3$ perpendicular to the normal vector $(3,-3,3)$. Denote by $n$ the normalized normal vector, i.e. $n=\frac{1}{\|(3,-3,3)\|}(3,-3,3)$.
Let $W$ be said plane, you can easily find an orthonormal basis $\left\{w_1,w_2\right\}$ of $W$. By definition $T(w_1)= w_1$ and $T(w_2)=w_2$. Also, by definition $T(n)=\frac{1}{2}n$.
Clearly $\alpha=\left\{n,w_1,w_2 \right\}$ is an orthonormal basis of $\mathbb{R}^3$ and the matrix of $T$ w.r.t. this basis is given by $$T_{\alpha}^{\alpha}=\begin{pmatrix}\frac{1}{2} &0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}.$$
This is fine, but it's probably not what you want as you want to express the matrix of $T$ w.r.t. the standard basis of $\mathbb{R}^3$. So all that's left to do is to do a base change which I hope you have seen in your linear algebra course at this point.
Hint: Given $v$, the half projection of $v$ lies exactly halfway between $v$ and the (full) projection of $v$.