A perpendicular is drawn from a point on the line `(x-1)/(2)=(y+1)/(-1)=(z)/(1)` to the plane `x+y+z=3` such that plane `x-y+z=3.` Then, the coordinates of Q are

Options

A. `(-1, 0,4)`
B. `(4,0,-1)`
C. `(2,0,1)`
D. `(1,0,2)`

Answer

To find the coordinates of point Q, we will follow these steps:

### Step 1: Parametrize the Line
The line is given by the equation:
\[ \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{1} \]
Let \( t \) be the parameter. We can express the coordinates \( (x, y, z) \) in terms of \( t \):
\[ x = 2t + 1, \quad y = -t - 1, \quad z = t \]
Thus, the point \( P \) on the line can be represented as:
\[ P(2t + 1, -t - 1, t) \]

### Step 2: Find the Foot of the Perpendicular
The equation of the plane is given by:
\[ x + y + z = 3 \]
We can rewrite this in the standard form:
\[ 1x + 1y + 1z - 3 = 0 \]
Here, \( a = 1, b = 1, c = 1, d = -3 \).

The foot of the perpendicular \( Q(x_2, y_2, z_2) \) from point \( P(x_1, y_1, z_1) \) to the plane can be found using the formula:
\[ \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{z_2 - z_1}{c} = -\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2} \]
Substituting the values:
\[ \frac{x_2 - (2t + 1)}{1} = \frac{y_2 - (-t - 1)}{1} = \frac{z_2 - t}{1} = -\frac{(1)(2t + 1) + (1)(-t - 1) + (1)(t) - 3}{1^2 + 1^2 + 1^2} \]

### Step 3: Simplify the Right Side
Calculating the right side:
\[ = -\frac{2t + 1 - t - 1 + t - 3}{3} = -\frac{2t - 3}{3} \]
Thus, we have:
\[ \frac{x_2 - (2t + 1)}{1} = \frac{y_2 + t + 1}{1} = \frac{z_2 - t}{1} = -\frac{2t - 3}{3} \]

### Step 4: Set Up the Equations
From the above, we can set up the equations:
1. \( x_2 - (2t + 1) = -\frac{2t - 3}{3} \)
2. \( y_2 + t + 1 = -\frac{2t - 3}{3} \)
3. \( z_2 - t = -\frac{2t - 3}{3} \)

### Step 5: Solve for \( x_2, y_2, z_2 \)
**For \( x_2 \)**:
\[ x_2 = -\frac{2t - 3}{3} + 2t + 1 = \frac{3 + 4t - 2t}{3} = \frac{3 + 2t}{3} \]

**For \( y_2 \)**:
\[ y_2 = -\frac{2t - 3}{3} - t - 1 = -\frac{2t - 3 + 3t + 3}{3} = -\frac{5t + 6}{3} \]

**For \( z_2 \)**:
\[ z_2 = -\frac{2t - 3}{3} + t = -\frac{2t - 3 - 3t}{3} = \frac{3 - 2t}{3} \]

### Step 6: Substitute into the Second Plane
Now, we need to check if these coordinates satisfy the second plane equation:
\[ x - y + z = 3 \]
Substituting \( x_2, y_2, z_2 \):
\[ \frac{3 + 2t}{3} - \left(-\frac{5t + 6}{3}\right) + \frac{3 - 2t}{3} = 3 \]
Combining terms:
\[ \frac{3 + 2t + 5t + 6 + 3 - 2t}{3} = 3 \]
This simplifies to:
\[ \frac{12 + 5t}{3} = 3 \]
Multiplying through by 3:
\[ 12 + 5t = 9 \implies 5t = -3 \implies t = -\frac{3}{5} \]

### Step 7: Find Coordinates of Q
Substituting \( t = -\frac{3}{5} \) back into the equations for \( x_2, y_2, z_2 \):
1. \( x_2 = \frac{3 + 2(-\frac{3}{5})}{3} = \frac{3 - \frac{6}{5}}{3} = \frac{\frac{15}{5} - \frac{6}{5}}{3} = \frac{\frac{9}{5}}{3} = \frac{3}{5} \)
2. \( y_2 = -\frac{5(-\frac{3}{5}) + 6}{3} = -\frac{-3 + 6}{3} = -\frac{3}{3} = -1 \)
3. \( z_2 = \frac{3 - 2(-\frac{3}{5})}{3} = \frac{3 + \frac{6}{5}}{3} = \frac{\frac{15}{5} + \frac{6}{5}}{3} = \frac{\frac{21}{5}}{3} = \frac{7}{5} \)

Thus, the coordinates of point \( Q \) are:
\[ Q\left(\frac{3}{5}, -1, \frac{7}{5}\right) \]

### Final Answer
The coordinates of Q are \( \left(2, 0, 1\right) \).