To find the coordinates of point Q, we will follow these steps systematically.
### Step 1: Identify the line and the planes
The line is given by the equation:
\[
\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{1}
\]
This can be parameterized as:
\[
x = 2r + 1, \quad y = -r - 1, \quad z = r
\]
where \( r \) is a parameter.
The two planes are:
1. Plane 1: \( x + y + z = 3 \)
2. Plane 2: \( x - y + z = 3 \)
### Step 2: Find the normal vector of the first plane
The normal vector \( \mathbf{n_1} \) of Plane 1 can be derived from the coefficients of \( x, y, z \):
\[
\mathbf{n_1} = (1, 1, 1)
\]
### Step 3: Find the coordinates of point P on the line
Using the parameterization, we can denote the coordinates of point P as:
\[
P(x_1, y_1, z_1) = (2r + 1, -r - 1, r)
\]
### Step 4: Use the formula for the foot of the perpendicular
The coordinates of point Q, which is the foot of the perpendicular from point P to Plane 1, can be found using the formula:
\[
\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{z_2 - z_1}{c} = -\frac{a x_1 + b y_1 + c z_1 + d}{a^2 + b^2 + c^2}
\]
where \( a, b, c \) are the coefficients of the plane equation \( ax + by + cz + d = 0 \).
For Plane 1:
- \( a = 1, b = 1, c = 1, d = -3 \)
### Step 5: Substitute the values into the formula
Substituting the values:
\[
\frac{x_2 - (2r + 1)}{1} = \frac{y_2 - (-r - 1)}{1} = \frac{z_2 - r}{1} = -\frac{(1)(2r + 1) + (1)(-r - 1) + (1)(r) - 3}{1^2 + 1^2 + 1^2}
\]
This simplifies to:
\[
\frac{x_2 - (2r + 1)}{1} = \frac{y_2 + r + 1}{1} = \frac{z_2 - r}{1} = -\frac{2r + 1 - r - 1 + r - 3}{3}
\]
\[
= -\frac{2r - 3}{3}
\]
### Step 6: Solve for \( x_2, y_2, z_2 \)
From the equations:
1. \( x_2 - (2r + 1) = -\frac{2r - 3}{3} \)
2. \( y_2 + r + 1 = -\frac{2r - 3}{3} \)
3. \( z_2 - r = -\frac{2r - 3}{3} \)
Solving these equations gives:
- For \( x_2 \):
\[
x_2 = -\frac{2r - 3}{3} + 2r + 1 = \frac{6 + 4r}{3}
\]
- For \( y_2 \):
\[
y_2 = -\frac{2r - 3}{3} - r - 1 = -\frac{5r + 3}{3}
\]
- For \( z_2 \):
\[
z_2 = -\frac{2r - 3}{3} + r = \frac{3 + r}{3}
\]
### Step 7: Substitute into the second plane equation
Now we substitute \( x_2, y_2, z_2 \) into the second plane equation \( x - y + z = 3 \):
\[
\frac{6 + 4r}{3} - \left(-\frac{5r + 3}{3}\right) + \frac{3 + r}{3} = 3
\]
Simplifying gives:
\[
\frac{6 + 4r + 5r + 3 + 3 + r}{3} = 3
\]
\[
\frac{12 + 10r}{3} = 3
\]
Multiplying through by 3:
\[
12 + 10r = 9 \implies 10r = -3 \implies r = -\frac{3}{10}
\]
### Step 8: Find the coordinates of Q
Substituting \( r = -\frac{3}{10} \) back into the equations for \( x_2, y_2, z_2 \):
1. \( x_2 = \frac{6 + 4(-\frac{3}{10})}{3} = \frac{6 - \frac{12}{10}}{3} = \frac{6 - 1.2}{3} = \frac{4.8}{3} = 1.6 \)
2. \( y_2 = -\frac{5(-\frac{3}{10}) + 3}{3} = \frac{15/10 + 30/10}{3} = \frac{45/10}{3} = \frac{15}{10} = 1.5 \)
3. \( z_2 = \frac{3 - \frac{3}{10}}{3} = \frac{30/10 - 3/10}{3} = \frac{27/10}{3} = \frac{9}{10} \)
Thus, the coordinates of point Q are:
\[
Q(1.6, 1.5, 0.9)
\]
