Given the points $\{p_0,p_1\}\in \Pi$ and the line $L\to p = p_2+\lambda \vec v$, determine a plane $\Pi$ parallel to $L$.
Characterizing the plane as $\Pi\to(p-p_0)\cdot\vec n=0$ we have the properties
$$
\cases{
(p_1-p_0)\cdot \vec n = 0\\
\vec v\cdot \vec n = 0\\
\|\vec n\|=1
}
$$
The last comes from normalization purposes.
Now as
$$
\cases{
p=(x,y,z)\\
p_0=(0,0,0)\\
p_1 = (1,2,3)\\
\vec v = (2,-1,1)
}
$$
we have to solve
$$
\cases{
(1,2,3)\cdot\vec n = 0\\
(2,-1,1)\cdot\vec n = 0\\
\|\vec n\|=1
}
$$
Actually you want to use the point $(2, -2, 1)$ to set up the equation, so you can use the line to solve it:
$$A(x-2) + B(y+2) + C(z-1) = 0$$
and then solve
$$A((1+2t)-2) + B((2-3t)+2) + C((-3+2t)-1) = 0$$
by plugging in some useful values of $t$. So
$$A(2t-1) + B(-3t+4) + C(2t-4) = 0$$
and try out $t=\frac{1}{2}$, then $t=\frac{4}{3}$, then $t=2$ and solve for $A,B,C$.
Best Answer
Here's another way (it's always good to know more than one way to solve a problem).
Your plane goes through $(-1,5,2)$ and $(2,7,3)$ (obtained by taking $t=0$ and $t=1$, respectively) and also $(2,4,-1)$. If your plane is $$ax+by+cz+d=0$$ then each of these three points gives you an equation relating the four unknowns $a,b,c,d$. So you get three homogeneous linear equations in 4 unknowns. Presumably, you know how to solve such a system. You'll get a one-parameter family of solutions (because $akx+bky+ckz+dk=0$ is the same plane, for any non-zero value of $k$); just pick any member of this family.