A plane bisects the line segment joining the points (1, 2, 3) and (-3, 4, 5) at right angles. Then this plane also passes through the point:

We begin with the two end-points of the segment: $$A(1,\,2,\,3) \text{ and } B(-3,\,4,\,5).$$ A plane that bisects this segment at right angles must satisfy two conditions:

1. It must pass through the mid-point of $$AB.$$ 2. Its normal vector must be parallel to the segment $$\overrightarrow{AB},$$ because the plane is perpendicular to the segment.

First, we find the mid-point. The mid-point formula is $$\left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2},\;\dfrac{z_1+z_2}{2}\right).$$ Substituting the coordinates of $$A$$ and $$B$$ we obtain

$$M\;=\;\left(\dfrac{1+(-3)}{2},\;\dfrac{2+4}{2},\;\dfrac{3+5}{2}\right) \;=\;\left(\dfrac{-2}{2},\;\dfrac{6}{2},\;\dfrac{8}{2}\right) \;=\;(-1,\,3,\,4).$$

Next, we determine a vector parallel to the segment. Using $$\overrightarrow{AB}=B-A,$$ we have

$$\overrightarrow{AB}=(-3-1,\;4-2,\;5-3)=(-4,\;2,\;2).$$

Because the plane is perpendicular to $$\overrightarrow{AB},$$ the vector $$\mathbf{n}=(-4,\,2,\,2)$$ serves as a normal vector to the required plane.

The point-normal form of a plane is stated as $$\mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0)=0,$$ where $$\mathbf{r}=(x,\,y,\,z)$$ is a general point on the plane, $$\mathbf{r}_0$$ is a known point on the plane (here $$M$$), and $$\mathbf{n}$$ is the normal vector. Substituting, we get

$$(-4,\,2,\,2)\cdot\bigl((x,\,y,\,z)-(-1,\,3,\,4)\bigr)=0.$$

This dot product expands to

$$-4\,(x+1)+2\,(y-3)+2\,(z-4)=0.$$

Now we open every bracket:

$$-4x-4+2y-6+2z-8=0.$$

Combining like terms,

$$-4x+2y+2z-18=0.$$

To simplify, we divide every term by $$2$$:

$$-2x+y+z-9=0.$$

Multiplying by $$-1$$ (merely changing the sign of each term) gives a neater form:

$$2x-y-z+9=0.$$

This is the required plane. To discover which option lies on it, we substitute the coordinates of each given point into the left-hand side $$2x-y-z+9$$ and check whether the result is zero.

Option A: $$(-3,\,2,\,1)$$ $$2(-3)-2-1+9=-6-2-1+9=0.$$ $$\checkmark$$

Option B: $$(3,\,2,\,1)$$ $$2(3)-2-1+9=6-3+9=12\neq0.$$

Option C: $$(1,\,2,\,-3)$$ $$2(1)-2+3+9=2-2+3+9=12\neq0.$$

Option D: $$(-1,\,2,\,3)$$ $$2(-1)-2-3+9=-2-2-3+9=2\neq0.$$

Only Option A satisfies the plane equation.

Hence, the correct answer is Option A.