Circle parameterization Draw the circle of radius two centered at (1,1, 1) and lying on the plane x + y+z = 3 Parameterize the circle. (Hint: Find two orthogonal unit vectors which are parallel to the plane.)
I don't need help with the drawing but so far I've found the unit vectors i, j, k using the plane equation, then the point (3,0,0)-(1,1,1), then crossed the two to find j.
Unfortunately, I don't know where to go from here and I'd appreciate some direction! (I am also unsure on whether I found the correct orthogonal unit vectors).

Let $C(1,1,1)$ be the center of the circle.

Fig. 1 : plane $x+y+z=3$ is materialized by triangle with vertices $(3,0,0),(0,3,0),(0,0,3).$
The normal vector to the plane is $U=(1,1,1)$. Let us take an arbitrary vector $V=(1,-1,0)$ orthogonal to $U$, here with norm $\sqrt{2}$ and a vector which is orthogonal to both $U$ and $V$, i.e., cross product $W:=U \times V=(1,1,-2)$ with norm $\sqrt{6}$. Then the circle can be characterized as the set of points $M$ such that (please note that we have normalized vectors $V$ and $W$):
$$M=C+2( \cos(t)\tfrac{1}{\sqrt{2}}V+\sin(t)\tfrac{1}{\sqrt{6}}W)\tag{1}$$
(the value $2$ in front of the parentheses accounts for the value of the radius).
(1) becomes, in coordinates :

$$\begin{cases}x&=&1+2 \cos(t)\tfrac{1}{\sqrt{2}}+2 \sin(t)\tfrac{1}{\sqrt{6}}\\
y&=&1-2 \cos(t)\tfrac{1}{\sqrt{2}}+2 \sin(t)\tfrac{1}{\sqrt{6}}\\
z&=&1- 4 \sin(t)\tfrac{1}{\sqrt{6}}\end{cases}$$

