Let $\pi$ be the plane with equation $x+y+z=0$, $A$ the point $(2,1,3)$, and $v$ the vector $(1,-2,1)$. Write the equation of the line $r$ passing through the point $B$, symmetric of $A$ with respect to $\pi$, parallel to $\pi$ and perpendicular to $v$.
My attempt: I have determined $B$ to be the point $(-2,-3,-1)$, I'm OK with this part.
Let $n = (1,1,1)$ be the normal vector to the plane $\pi$ and $u = (a,b,c)$ the direction vector of $r$. Then we have $$<u, n> = <u, v> = 0$$ so $a+b+c = 0$ and $a-2b+c = 0$.
So it seems that any vector $u = (\alpha, 0, -\alpha)$ will make its resulting line passing through $B$ satisfy the conditions, so there isn't one line as the problem states, but infinitely many. Or am I missing something in the situation?
Let $AB\cap\pi=\{C\}$.
Thus, $$C(2+t,1+t,3+t),$$ which gives $$2+t+1+t+3+t=0,$$ $$t=-2$$ and $$C(0,-1,1)$$ and since $B$ is a mid-point of $AB$, we obtain: $$B(-2,-3,-1).$$ Also, let $\vec{l}(a,b,c)$ is parallel to needed line.
Thus, $$a+b+c=0$$ and $$a-2b+c=0,$$ which gives $$b=0,$$ $$c=-a$$ and after assuming $a=1$ we obtain: $$\vec{l}(1,0,-1)$$ and we got for needed line: $$\{(-2+t,-3,-1-t)|t\in\mathbb R\}.$$ Now, we see that you got the same result.
It's an unique line, of course.