So, you are given the vector $(2,1,-3)$. Let $(2k,k,-3k)$ be the orthogonal projection of $(1,1,1)$ on $(2,1,-3)$. Then, $(2k-1,k-1,-3k-1)$ and $(2,1,-3)$ are orthogonal, giving:
$4k-2+k-1+9k+3=0$ i.e. $k=0$. So, $(0,0,0)$, the origin is the projection. Hence, the line contains the points $(0,0,0)$ and $(1,1,1)$, so its equation is $x=y=z$, if my calculations are correct!
If $\mathbf{n}$ is the normal vector to the given plane and $\mathbf{p}$ is the point through which the line is supposed to pass, then the equation of the line will be of the form $\mathbf{r}=\mathbf{p}+t\mathbf{n}$. You already have both those vectors so you don't need anything else.
Best Answer
Let $\vec x=\vec a+\lambda\vec u$ be the equation of the line and $\vec p$ be any vector. Then the orthogonal projection of $\vec p-\vec a$ along $\vec u$ is $\dfrac{\langle \vec p-\vec a,\vec u\rangle}{\|u\|^2}\vec u$. Can you finish from here?