Find the point of intersection of line passing through `(0,0,1)` and the intersection lines `x+2u+z=1,-x+y-2z and x+y=2,x+z=2` with the `x y` plane.

Options

A. `((5)/(3),-(1)/(3),0)`
B. (1,1,0)
C. `((2)/(3),-(1)/(3),0)`
D. `(-(5)/(3),(1)/(3),0)`

Explanation

To find the point of intersection of the line passing through the point (0, 0, 1) and the intersection of the given planes with the xy-plane, we can follow these steps:

### Step 1: Write the equations of the planes
The equations of the planes are given as:
1. \( x + 2y + z = 1 \) (Plane 1)
2. \( -x + y - 2z = 0 \) (Plane 2)
3. \( x + y = 2 \) (Plane 3)
4. \( x + z = 2 \) (Plane 4)

### Step 2: Find the intersection line of the planes
To find the line of intersection of the planes, we can solve the equations simultaneously.

First, we can express one variable in terms of the others. Let's solve the first two equations:

From Plane 1:
\[ z = 1 - x - 2y \tag{1} \]

From Plane 2:
\[ y = x + 2z \tag{2} \]

Substituting (1) into (2):
\[ y = x + 2(1 - x - 2y) \]
\[ y = x + 2 - 2x - 4y \]
\[ 5y = 2 - x \]
\[ y = \frac{2 - x}{5} \tag{3} \]

### Step 3: Substitute \(y\) into Plane 3
Now substitute (3) into Plane 3:
\[ x + \frac{2 - x}{5} = 2 \]
Multiplying through by 5 to eliminate the fraction:
\[ 5x + 2 - x = 10 \]
\[ 4x = 8 \]
\[ x = 2 \]

### Step 4: Find \(y\) and \(z\)
Now substitute \(x = 2\) back into (3) to find \(y\):
\[ y = \frac{2 - 2}{5} = 0 \]

Now substitute \(x = 2\) and \(y = 0\) into (1) to find \(z\):
\[ z = 1 - 2 - 2(0) = -1 \]

### Step 5: Find the point of intersection with the xy-plane
The point of intersection of the line with the xy-plane occurs when \(z = 0\). We need to find the values of \(x\) and \(y\) when \(z = 0\).

Using Plane 1:
\[ x + 2y + 0 = 1 \implies x + 2y = 1 \tag{4} \]

Using Plane 2:
\[ -x + y - 0 = 2 \implies -x + y = 2 \implies y = x + 2 \tag{5} \]

### Step 6: Solve equations (4) and (5)
Substituting (5) into (4):
\[ x + 2(x + 2) = 1 \]
\[ x + 2x + 4 = 1 \]
\[ 3x + 4 = 1 \]
\[ 3x = -3 \implies x = -1 \]

Now substitute \(x = -1\) back into (5):
\[ y = -1 + 2 = 1 \]

### Final Step: Write the point of intersection
Thus, the point of intersection of the line with the xy-plane is:
\[ (-1, 1, 0) \]

### Summary of the Solution
The point of intersection of the line passing through (0, 0, 1) and the intersection of the given planes with the xy-plane is:
\[ \boxed{(-1, 1, 0)} \]