If P is the point (2,1,6) find the point Q such that PQ is perpendiculr to the plane in x+y-2z =3 and the mid point of PQ lies on it.

Answer

To solve the problem, we need to find the point \( Q \) such that the line segment \( PQ \) is perpendicular to the plane defined by the equation \( x + y - 2z = 3 \) and the midpoint of \( PQ \) lies on this plane.

### Step 1: Identify the normal vector of the plane
The equation of the plane is given by:
\[ x + y - 2z = 3 \]
The normal vector \( \mathbf{n} \) to the plane can be extracted from the coefficients of \( x, y, \) and \( z \). Thus, the normal vector is:
\[ \mathbf{n} = (1, 1, -2) \]

### Step 2: Determine the direction of line segment \( PQ \)
Since \( PQ \) is perpendicular to the plane, the direction of \( PQ \) will be along the normal vector \( \mathbf{n} \). Therefore, we can express the coordinates of point \( Q \) in terms of a parameter \( \lambda \) as follows:
Let \( P(2, 1, 6) \) be the coordinates of point \( P \). Then, the coordinates of point \( Q \) can be expressed as:
\[ Q(x, y, z) = P + \lambda \cdot \mathbf{n} = (2 + \lambda, 1 + \lambda, 6 - 2\lambda) \]

### Step 3: Find the midpoint of segment \( PQ \)
The midpoint \( M \) of segment \( PQ \) is given by:
\[ M = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}, \frac{z_P + z_Q}{2} \right) \]
Substituting the coordinates of \( P \) and \( Q \):
\[ M = \left( \frac{2 + (2 + \lambda)}{2}, \frac{1 + (1 + \lambda)}{2}, \frac{6 + (6 - 2\lambda)}{2} \right) \]
This simplifies to:
\[ M = \left( \frac{4 + \lambda}{2}, \frac{2 + \lambda}{2}, \frac{12 - 2\lambda}{2} \right) = \left( 2 + \frac{\lambda}{2}, 1 + \frac{\lambda}{2}, 6 - \lambda \right) \]

### Step 4: Substitute the midpoint into the plane equation
Since the midpoint \( M \) lies on the plane, we substitute its coordinates into the plane equation:
\[ \left( 2 + \frac{\lambda}{2} \right) + \left( 1 + \frac{\lambda}{2} \right) - 2\left( 6 - \lambda \right) = 3 \]
Simplifying this:
\[ 2 + \frac{\lambda}{2} + 1 + \frac{\lambda}{2} - 12 + 2\lambda = 3 \]
Combining like terms gives:
\[ 3 + 3\lambda - 12 = 3 \]
This simplifies to:
\[ 3\lambda - 9 = 0 \implies 3\lambda = 9 \implies \lambda = 3 \]

### Step 5: Find the coordinates of point \( Q \)
Now, substituting \( \lambda = 3 \) back into the expression for \( Q \):
\[ Q = (2 + 3, 1 + 3, 6 - 2 \cdot 3) = (5, 4, 0) \]

### Final Answer
Thus, the coordinates of point \( Q \) are:
\[ \boxed{(5, 4, 0)} \]