On which of the following lines lies the point of intersection of the line, `(x-4)/(2)=(y-5)/(2)=(z-3)/(1)` and the plane, `x+y+z=2`?

Options

A. `(x-4)/(2) = (y-5)/(2) = (z-5)/(-1)`
B. `(x+3)/(3) = (4-y)/(3) = (z+1)/(-2)`
C. `(x-2)/(2) = (y-3)/(2) = (z+3)/(3)`
D. `(x-1)/(1) = (y-3)/(2) = (z+4)/(-5)`

Answer

To find the point of intersection of the line given by the equation \((x-4)/(2)=(y-5)/(2)=(z-3)/(1)\) and the plane defined by the equation \(x+y+z=2\), we can follow these steps:

### Step 1: Parametrize the Line
The line can be expressed in parametric form. Let \(\lambda\) be the parameter:
- From \((x-4)/2 = \lambda\), we have \(x = 2\lambda + 4\).
- From \((y-5)/2 = \lambda\), we have \(y = 2\lambda + 5\).
- From \((z-3)/1 = \lambda\), we have \(z = \lambda + 3\).

Thus, the parametric equations of the line are:
\[ x = 2\lambda + 4, \quad y = 2\lambda + 5, \quad z = \lambda + 3 \]

### Step 2: Substitute into the Plane Equation
Now we substitute \(x\), \(y\), and \(z\) into the plane equation \(x + y + z = 2\):
\[ (2\lambda + 4) + (2\lambda + 5) + (\lambda + 3) = 2 \]

### Step 3: Simplify the Equation
Combining like terms:
\[ 2\lambda + 4 + 2\lambda + 5 + \lambda + 3 = 2 \]
\[ 5\lambda + 12 = 2 \]

### Step 4: Solve for \(\lambda\)
Now, isolate \(\lambda\):
\[ 5\lambda = 2 - 12 \]
\[ 5\lambda = -10 \]
\[ \lambda = -2 \]

### Step 5: Find the Point of Intersection
Substituting \(\lambda = -2\) back into the parametric equations:
- For \(x\):
\[ x = 2(-2) + 4 = -4 + 4 = 0 \]
- For \(y\):
\[ y = 2(-2) + 5 = -4 + 5 = 1 \]
- For \(z\):
\[ z = -2 + 3 = 1 \]

Thus, the point of intersection is \((0, 1, 1)\).

### Step 6: Check Which Line Contains the Point
Now we need to check which of the given lines contains the point \((0, 1, 1)\). We will substitute the coordinates into each line's equation.

1. **Option A**: Check if \((0, 1, 1)\) satisfies the line equation.
\[ \frac{0 - 4}{2} = \frac{1 - 5}{2} = \frac{1 - 3}{1} \]
This simplifies to:
\[ -2 \neq -2 \neq -2 \quad \text{(not equal)} \]

2. **Option B**: Check if \((0, 1, 1)\) satisfies the line equation.
\[ \frac{0 + 3}{3} = \frac{4 - 1}{3} = \frac{1 + 1}{-2} \]
This simplifies to:
\[ 1 = 1 \neq -1 \quad \text{(not equal)} \]

3. **Option C**: Check if \((0, 1, 1)\) satisfies the line equation.
\[ \frac{0 - 2}{2} = \frac{1 - 3}{2} = \frac{1 + 3}{3} \]
This simplifies to:
\[ -1 \neq -1 \neq \frac{4}{3} \quad \text{(not equal)} \]

4. **Option D**: Check if \((0, 1, 1)\) satisfies the line equation.
\[ \frac{0 - 1}{1} = \frac{1 - 3}{2} = \frac{1 + 4}{-5} \]
This simplifies to:
\[ -1 = -1 = -1 \quad \text{(all equal)} \]

### Conclusion
The point of intersection \((0, 1, 1)\) lies on line D.