The plane which bisects the line segment joining the points $$(-3, -3, 4)$$ and $$(3, 7, 6)$$ at right angles, passes through which one of the following points?

We have two given points $$A(-3,\,-3,\,4)$$ and $$B(3,\,7,\,6)$$. The plane we are looking for must satisfy two geometrical facts: it must pass through the midpoint of $$AB$$ (so that it “bisects” the segment) and it must be perpendicular to $$AB$$ (so that it “bisects at right angles”).

First we find the direction vector of the segment $$AB$$. Using the section formula for a vector between two points,

$$\overrightarrow{AB}=B-A=(3-(-3),\,7-(-3),\,6-4)=(6,\,10,\,2).$$

Next we find the midpoint $$M$$ of the segment $$AB$$. The midpoint formula in 3D is

$$M=\left(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2},\,\dfrac{z_1+z_2}{2}\right).$$

Substituting $$A(-3,\,-3,\,4)$$ and $$B(3,\,7,\,6)$$, we get

$$M=\left(\dfrac{-3+3}{2},\,\dfrac{-3+7}{2},\,\dfrac{4+6}{2}\right)=(0,\,2,\,5).$$

Because the required plane is perpendicular to $$AB$$, the normal vector $$\mathbf{n}$$ of the plane is the same as $$\overrightarrow{AB}$$, namely $$\mathbf{n}=(6,\,10,\,2).$$

The point-normal form of a plane is stated as follows:

$$\mathbf{n}\cdot\bigl(\mathbf{r}-\mathbf{r_0}\bigr)=0,$$

where $$\mathbf{n}$$ is a normal vector, $$\mathbf{r}=(x,\,y,\,z)$$ an arbitrary point on the plane, and $$\mathbf{r_0}$$ a fixed point on the plane (here, the midpoint $$M$$).

Taking $$\mathbf{n}=(6,\,10,\,2)$$ and $$\mathbf{r_0}=M(0,\,2,\,5)$$, we write

$$6\bigl(x-0\bigr)+10\bigl(y-2\bigr)+2\bigl(z-5\bigr)=0.$$

Now we expand each term step by step:

$$6x \;+\;10y-20 \;+\;2z-10 =0.$$

Combining the constant terms $$-20-10=-30$$, we obtain

$$6x+10y+2z-30=0.$$

For convenience we divide every term by $$2$$ (this does not change the set of solutions) to get a simpler but equivalent equation:

$$3x+5y+z-15=0.$$

Any point that satisfies $$3x+5y+z-15=0$$ lies on the required plane. We now test each option.

Option A: $$(2,1,3)$$

$$3(2)\;+\;5(1)\;+\;3\;-\;15=6+5+3-15=-1\neq0$$ so this point is not on the plane.

Option B: $$(4,1,-2)$$

$$3(4)\;+\;5(1)\;+\;(-2)\;-\;15=12+5-2-15=0$$ so this point is on the plane.

Option C: $$(4,-1,7)$$

$$3(4)\;+\;5(-1)\;+\;7\;-\;15=12-5+7-15=-1\neq0$$ so this point is not on the plane.

Option D: $$(-2,3,5)$$

$$3(-2)\;+\;5(3)\;+\;5\;-\;15=-6+15+5-15=-1\neq0$$ so this point is not on the plane.

Only Option B satisfies the equation of the plane. Hence, the correct answer is Option B.