Q is the image of point P(1, -2, 3) with respect to the plane `x-y+z=7`. The distance of Q from the origin is.

Options

A. `sqrt((70)/(3))`
B. `(1)/(2) sqrt((70)/(3))`
C. `sqrt((35)/(3))`
D. `sqrt((15)/(2))`

Answer

To find the distance of point Q (the image of point P with respect to the plane \(x - y + z = 7\)) from the origin, we will follow these steps:

### Step 1: Identify the given point and the plane
The point \(P\) is given as \(P(1, -2, 3)\) and the equation of the plane is \(x - y + z = 7\).

### Step 2: Find the normal vector of the plane
The normal vector \(\mathbf{n}\) of the plane \(x - y + z = 7\) can be derived from the coefficients of \(x\), \(y\), and \(z\) in the equation. Thus, the normal vector is:
\[ \mathbf{n} = (1, -1, 1) \]

### Step 3: Find the foot of the perpendicular from point P to the plane
To find the foot of the perpendicular \(M\) from point \(P\) to the plane, we can use the parametric equations of the line that passes through \(P\) in the direction of the normal vector \(\mathbf{n}\):
\[ \frac{x - 1}{1} = \frac{y + 2}{-1} = \frac{z - 3}{1} = \lambda \]
This gives us the parametric equations:
\[ x = 1 + \lambda, \quad y = -2 - \lambda, \quad z = 3 + \lambda \]

### Step 4: Substitute the parametric equations into the plane equation
Substituting these into the plane equation \(x - y + z = 7\):
\[ (1 + \lambda) - (-2 - \lambda) + (3 + \lambda) = 7 \]
Simplifying this:
\[ 1 + \lambda + 2 + \lambda + 3 + \lambda = 7 \implies 3\lambda + 6 = 7 \implies 3\lambda = 1 \implies \lambda = \frac{1}{3} \]

### Step 5: Find the coordinates of point M
Substituting \(\lambda = \frac{1}{3}\) back into the parametric equations:
\[ x_M = 1 + \frac{1}{3} = \frac{4}{3}, \quad y_M = -2 - \frac{1}{3} = -\frac{7}{3}, \quad z_M = 3 + \frac{1}{3} = \frac{10}{3} \]
Thus, the coordinates of point \(M\) are:
\[ M\left(\frac{4}{3}, -\frac{7}{3}, \frac{10}{3}\right) \]

### Step 6: Find the coordinates of point Q using the midpoint formula
Using the midpoint formula, where \(M\) is the midpoint of segment \(PQ\):
\[ \left(\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}, \frac{z_P + z_Q}{2}\right) = \left(\frac{4}{3}, -\frac{7}{3}, \frac{10}{3}\right) \]
This gives us three equations:
1. \(\frac{1 + x_Q}{2} = \frac{4}{3}\)
2. \(\frac{-2 + y_Q}{2} = -\frac{7}{3}\)
3. \(\frac{3 + z_Q}{2} = \frac{10}{3}\)

Solving these equations:
1. From \(\frac{1 + x_Q}{2} = \frac{4}{3}\):
\[ 1 + x_Q = \frac{8}{3} \implies x_Q = \frac{8}{3} - 1 = \frac{5}{3} \]
2. From \(\frac{-2 + y_Q}{2} = -\frac{7}{3}\):
\[ -2 + y_Q = -\frac{14}{3} \implies y_Q = -\frac{14}{3} + 2 = -\frac{14}{3} + \frac{6}{3} = -\frac{8}{3} \]
3. From \(\frac{3 + z_Q}{2} = \frac{10}{3}\):
\[ 3 + z_Q = \frac{20}{3} \implies z_Q = \frac{20}{3} - 3 = \frac{20}{3} - \frac{9}{3} = \frac{11}{3} \]

Thus, the coordinates of point \(Q\) are:
\[ Q\left(\frac{5}{3}, -\frac{8}{3}, \frac{11}{3}\right) \]

### Step 7: Calculate the distance of point Q from the origin
Using the distance formula:
\[ d = \sqrt{x_Q^2 + y_Q^2 + z_Q^2} \]
Substituting the coordinates of \(Q\):
\[ d = \sqrt{\left(\frac{5}{3}\right)^2 + \left(-\frac{8}{3}\right)^2 + \left(\frac{11}{3}\right)^2} \]
Calculating each term:
\[ = \sqrt{\frac{25}{9} + \frac{64}{9} + \frac{121}{9}} = \sqrt{\frac{210}{9}} = \frac{\sqrt{210}}{3} \]

### Final Answer
The distance of point \(Q\) from the origin is:
\[ \frac{\sqrt{210}}{3} \text{ units} \]