Finding the Image of a Point in a Plane

Given three points:

\( A(2,1,0), B(4,1,1), C(5,0,1) \)

The plane equation passing through these points is found using the normal vector.

Step 1: Compute Normal Vector

Vectors:

\[ \overrightarrow{AB} = (2, 0, 1), \quad \overrightarrow{AC} = (3, -1, 1) \]

Normal vector \( \mathbf{N} \) is given by:

\[ \mathbf{N} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 1 \\ 3 & -1 & 1 \end{vmatrix} \]

Solving, we get \( (1,1,-2) \), so the equation of the plane is:

\[ x + y - 2z = 3 \]

Step 2: Find Foot of Perpendicular

Point \( R(2,1,6) \) is reflected about the plane. The perpendicular meets the plane at:

\[ x_0 = 2 + \lambda(1), \quad y_0 = 1 + \lambda(1), \quad z_0 = 6 + \lambda(-2) \]

Substituting into the plane equation:

\[ (2 + \lambda) + (1 + \lambda) - 2(6 - 2\lambda) = 3 \]

Solving for \( \lambda \), we get \( \lambda = 2 \) and thus:

\( Q(4,3,2) \).

Step 3: Compute Image

Using the formula:

\[ x' = 2x_0 - x, \quad y' = 2y_0 - y, \quad z' = 2z_0 - z \]

We obtain:

\( R'(6,5,-2) \).

Final Answer:

\[ \mathbf{(6,5,-2)} \]