Which of the following lines lie on the plane `x+2y-z+4=0?` a. `(x-1)/1=y/(-1)=(z-5)/1` b. `x-y+z=2x+y-z=0` c. ` hat r=2 hat i- hat j+4 hat k+lambda(3 hat i+ hat j+5 hat k)` d. none of these

Options

A. `(x-1)/(1)=(y)/(-1)=(z-5)/(-1)`
B. `x-y+z=2x+y-z=0`
C. `vecr=2hati-hatj+4hatk+lamda(3hati+hatj+5hatk)`
D. none of these

Explanation

To determine which of the given lines lie on the plane defined by the equation \( x + 2y - z + 4 = 0 \), we will analyze each option step by step.

### Step 1: Identify the normal vector of the plane
The equation of the plane can be written as:
\[ x + 2y - z + 4 = 0 \]
From this equation, we can identify the normal vector \( \mathbf{n} \) of the plane as:
\[ \mathbf{n} = (1, 2, -1) \]

### Step 2: Analyze each option

#### Option A:
\[ \frac{x-1}{1} = \frac{y}{-1} = \frac{z-5}{1} \]
This can be rewritten in parametric form:
- Let \( t = \frac{x-1}{1} \) then:
- \( x = t + 1 \)
- \( y = -t \)
- \( z = t + 5 \)

Substituting these into the plane equation:
\[ (t + 1) + 2(-t) - (t + 5) + 4 = 0 \]
Simplifying:
\[ t + 1 - 2t - t - 5 + 4 = 0 \]
\[ 0 = 0 \]
This holds true, hence **Option A lies on the plane**.

#### Option B:
\[ x - y + z = 2 \quad \text{and} \quad 2x + y - z = 0 \]
We can find the direction ratios of the lines represented by these equations. The direction ratios can be derived from the coefficients of \( x, y, z \).

For the first equation:
\[ x - y + z - 2 = 0 \implies (1, -1, 1) \]
For the second equation:
\[ 2x + y - z = 0 \implies (2, 1, -1) \]
Now, we need to check if these direction ratios satisfy the plane equation.

Taking a point on the line, we can use \( x = 0, y = 0 \) in the first equation:
\[ 0 - 0 + z = 2 \implies z = 2 \implies (0, 0, 2) \]
Substituting into the plane equation:
\[ 0 + 2(0) - 2 + 4 = 0 \implies 2 \neq 0 \]
Thus, **Option B does not lie on the plane**.

#### Option C:
\[ \mathbf{r} = 2\mathbf{i} - \mathbf{j} + 4\mathbf{k} + \lambda(3\mathbf{i} + \mathbf{j} + 5\mathbf{k}) \]
This can be expressed in parametric form:
- For \( \lambda = 0 \):
- \( (x, y, z) = (2, -1, 4) \)
- For \( \lambda = 1 \):
- \( x = 2 + 3\lambda \)
- \( y = -1 + \lambda \)
- \( z = 4 + 5\lambda \)

Substituting \( (2, -1, 4) \) into the plane equation:
\[ 2 + 2(-1) - 4 + 4 = 0 \implies 0 = 0 \]
This holds true, hence **Option C lies on the plane**.

### Conclusion:
The lines that lie on the plane \( x + 2y - z + 4 = 0 \) are:
- Option A
- Option C