Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is

We are asked to find the mirror-image of the point $$R(2,1,6)$$ in the plane that passes through the three given points $$A(2,1,0),\;B(4,1,1),\;C(5,0,1).$$ In other words, we must first write the equation of the plane, then locate the foot of the perpendicular from $$R$$ to this plane, and finally obtain the reflected point.

First we construct two independent direction vectors that lie on the plane:

$$\overrightarrow{AB}=B-A=(4-2,\;1-1,\;1-0)=(2,0,1),$$

$$\overrightarrow{AC}=C-A=(5-2,\;0-1,\;1-0)=(3,-1,1).$$

The normal vector $$\vec n$$ to the plane is obtained by the cross product of these two vectors:

$$ \vec n=\overrightarrow{AB}\times\overrightarrow{AC} =\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 2 & 0 & 1\\ 3 & -1 & 1 \end{vmatrix} =\mathbf i(0\cdot1-1\cdot(-1)) -\mathbf j(2\cdot1-1\cdot3) +\mathbf k(2\cdot(-1)-0\cdot3) =(1,1,-2). $$

Thus $$\vec n=(1,1,-2).$$ Using point $$A(2,1,0)$$ we write the plane in point-normal form $$\vec n\cdot\bigl((x,y,z)-(2,1,0)\bigr)=0$$:

$$1(x-2)+1(y-1)-2(z-0)=0,$$

which simplifies to

$$x+y-2z-3=0.$$

Now we evaluate this plane expression at the point $$R(2,1,6)$$ :

$$ S_R=x+y-2z-3 =2+1-2\cdot6-3 =3-12-3 =-12. $$

The formula for the foot of the perpendicular $$H$$ from a point $$R$$ to a plane $$ax+by+cz+d=0$$ is

$$ \vec{OH}=\vec{OR}-\frac{a x_R+b y_R+c z_R+d}{a^2+b^2+c^2}\,(a,b,c), $$

where $$(a,b,c)$$ is the normal vector. Here $$a=1,\;b=1,\;c=-2,$$ and $$d=-3.$$ Hence

$$ \frac{S_R}{a^2+b^2+c^2} =\frac{-12}{1^2+1^2+(-2)^2} =\frac{-12}{6} =-2. $$

So

$$ \vec{OH} =\vec{OR}-(-2)\,(1,1,-2) =\vec{OR}+(2,2,-4). $$

Writing the coordinates explicitly,

$$ H=(2,1,6)+(2,2,-4)=(4,3,2). $$

This point $$H(4,3,2)$$ is the foot of the perpendicular. To obtain the image $$R'$$ of $$R$$ in the plane, we move from $$R$$ to $$H$$ and then the same distance again beyond the plane:

$$ \overrightarrow{HR}=R-H=(2,1,6)-(4,3,2)=(-2,-2,4), $$

so

$$ R'=H+\overrightarrow{HR} =H-( -2,-2,4) =H+(2,2,-4) =(4+2,\;3+2,\;2-4) =(6,5,-2). $$

We have obtained $$R'(6,5,-2),$$ which matches Option B.

Hence, the correct answer is Option B.