Consider the line $$L$$ given by the equation $$\frac{x-3}{2} = \frac{y-1}{1} = \frac{z-2}{1}$$. Let $$Q$$ be the mirror image of the point $$(2, 3, -1)$$ with respect to $$L$$. Let a plane $$P$$ be such that it passes through $$Q$$, and the line $$L$$ is perpendicular to $$P$$. Then which of the following points is on the plane $$P$$?

Given line

$$L:\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$$

Hence a point on the line is

$$A(3,1,2)$$

and direction vector is

$$\vec d=(2,1,1)$$

Given point

$$P(2,3,-1)$$

Let the foot of perpendicular from $$P$$ to line $$L$$ be

$$R(3+2t,1+t,2+t)$$

Since

$$\overrightarrow{PR}\perp \vec d,$$

we have

$$[(3+2t)-2,\ (1+t)-3,\ (2+t)+1]\cdot(2,1,1)=0$$

$$(1+2t,\ t-2,\ t+3)\cdot(2,1,1)=0$$

$$2(1+2t)+(t-2)+(t+3)=0$$

$$2+4t+t-2+t+3=0$$

$$6t+3=0$$

$$t=-\frac12$$

Therefore,

$$R=\left(3-1,\ 1-\frac12,\ 2-\frac12\right)$$

$$=\left(2,\frac12,\frac32\right)$$

Now $$Q$$ is the mirror image of $$P$$ about line $$L,$$ so

$$R$$

is midpoint of

$$PQ$$

Hence,

$$Q=2R-P$$

$$=\left(4,1,3\right)-(2,3,-1)$$

$$=(2,-2,4)$$

Now plane $$P$$ passes through $$Q$$ and is perpendicular to line $$L.$$

Therefore normal vector of plane is

$$\vec n=(2,1,1)$$

Equation of plane:

$$2(x-2)+(y+2)+(z-4)=0$$

$$2x+y+z-6=0$$

Hence any point satisfying

$$2x+y+z=6$$

lies on the plane.

Therefore the required point $$(1, 2, 2)$$