The mirror image of the point (1, 2, 3), in a plane is $$\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$$. Which of the following points lies on this plane?

We denote the given point by $$P(1,\,2,\,3)$$ and its mirror image by $$P'\!\left(-\dfrac73,\,-\dfrac43,\,-\dfrac13\right).$$ By definition of reflection, the unknown plane must satisfy two facts: it is perpendicular to the segment $$PP'$$ and it passes through the midpoint of that segment.

First we find the vector $$\overrightarrow{PP'}.$$ Subtracting the coordinates we get

$$\overrightarrow{PP'} = \left(-\dfrac73-1,\;-\dfrac43-2,\;-\dfrac13-3\right) = \left(-\dfrac{10}3,\;-\dfrac{10}3,\;-\dfrac{10}3\right).$$

This vector is a common multiple of $$(-1,\,-1,\,-1),$$ so the normal vector of the plane may be taken as $$\mathbf n=(1,\,1,\,1).$$

Now we need the midpoint $$M$$ of $$PP'$$. Using the midpoint formula

$$M\! \left(\dfrac{1+\left(-\dfrac73\right)}2,\; \dfrac{2+\left(-\dfrac43\right)}2,\; \dfrac{3+\left(-\dfrac13\right)}2\right)$$

we compute each coordinate step by step:

$$\dfrac{1+\left(-\dfrac73\right)}2 = \dfrac{\dfrac33-\dfrac73}{2} = \dfrac{-\dfrac43}{2} = -\dfrac23,$$

$$\dfrac{2+\left(-\dfrac43\right)}2 = \dfrac{\dfrac63-\dfrac43}{2} = \dfrac{\dfrac23}{2} = \dfrac13,$$

$$\dfrac{3+\left(-\dfrac13\right)}2 = \dfrac{\dfrac93-\dfrac13}{2} = \dfrac{\dfrac83}{2} = \dfrac43.$$

So $$M\!\left(-\dfrac23,\,\dfrac13,\,\dfrac43\right).$$

Because $$\mathbf n=(1,\,1,\,1)$$ is normal to the plane and the plane passes through $$M,$$ its equation is written (stating the point-normal form first)

$$\mathbf n\cdot\bigl((x,y,z)-M\bigr)=0,$$ i.e. $$1\,(x+ \tfrac23)\;+\;1\,(y-\tfrac13)\;+\;1\,(z-\tfrac43)=0.$$

Adding the three terms gives

$$x+\dfrac23+y-\dfrac13+z-\dfrac43=0.$$

Combining the constants,

$$x+y+z+\left(\dfrac23-\dfrac13-\dfrac43\right)=0 \;\;\Longrightarrow\;\; x+y+z-1=0.$$

Thus the plane is $$x+y+z=1.$$

We now test each option:

A. For $$(1,1,1)$$ we find $$1+1+1=3\neq1,$$ so it does not lie on the plane.

B. For $$(1,-1,1)$$ we have $$1+(-1)+1=1,$$ so this point satisfies the equation.

C. For $$(-1,-1,1)$$ we get $$-1-1+1=-1\neq1,$$ hence not on the plane.

D. For $$(-1,-1,-1)$$ we have $$-1-1-1=-3\neq1,$$ so it is also not on the plane.

Only option B fulfills the condition.

Hence, the correct answer is Option B.