The equation of the line which intersect each of the two lines `2x+y-1 =0 =x-2y+3z` and `3x -y+z+2 = 0 = 4x + 5y - 2z-3 = 0` and is parallel to `(x)/(1)=(y)/(2)=(z)/(3)` is

Options

A. `4x+7y-6z-1=0=2x-7y+4z+3`
B. `4x+7y-6z-4=0=2x-7y+4z+2`
C. `4x+7y-6z-3=0=2x-7y+4z+7`
D. `4x+7y-6z+7=0=2x-7y+4z-3`

Answer

To find the equation of the line that intersects the two given lines and is parallel to the line defined by \(\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\), we can follow these steps:

### Step 1: Identify the given lines
The first line is defined by the equations:
1. \(2x + y - 1 = 0\)
2. \(x - 2y + 3z = 0\)

The second line is defined by the equations:
1. \(3x - y + z + 2 = 0\)
2. \(4x + 5y - 2z - 3 = 0\)

### Step 2: Form the planes from the lines
We can form two planes from the equations of the lines.

**For the first plane:**
\[ 2x + y - 1 + \lambda (x - 2y + 3z) = 0 \]
This expands to:
\[ (2 + \lambda)x + (1 - 2\lambda)y + (3\lambda)z - 1 = 0 \]

**For the second plane:**
\[ 3x - y + z + 2 + \mu (4x + 5y - 2z - 3) = 0 \]
This expands to:
\[ (3 + 4\mu)x + (-1 + 5\mu)y + (1 - 2\mu)z + (2 - 3\mu) = 0 \]

### Step 3: Find the conditions for parallelism
The line we are looking for is parallel to the line defined by:
\[ \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \]
This means the direction ratios of the line are \( (1, 2, 3) \).

For the line to be parallel to this, the normal vectors of the planes must satisfy:
\[ (2 + \lambda) \cdot 1 + (1 - 2\lambda) \cdot 2 + (3\lambda) \cdot 3 = 0 \]
This simplifies to:
\[ 2 + \lambda + 2 - 4\lambda + 9\lambda = 0 \]
\[ 6\lambda + 4 = 0 \implies \lambda = -\frac{2}{3} \]

For the second plane:
\[ (3 + 4\mu) \cdot 1 + (-1 + 5\mu) \cdot 2 + (1 - 2\mu) \cdot 3 = 0 \]
This simplifies to:
\[ 3 + 4\mu - 2 + 10\mu + 3 - 6\mu = 0 \]
\[ 8\mu + 4 = 0 \implies \mu = -\frac{1}{2} \]

### Step 4: Substitute back to find the equations of the planes
**First Plane:**
Substituting \(\lambda = -\frac{2}{3}\):
\[ (2 - \frac{2}{3})x + (1 + \frac{4}{3})y + (-2)z - 1 = 0 \]
This simplifies to:
\[ \frac{4}{3}x + \frac{7}{3}y - 2z - 1 = 0 \implies 4x + 7y - 6z - 3 = 0 \]

**Second Plane:**
Substituting \(\mu = -\frac{1}{2}\):
\[ (3 - 2)x + (-1 - \frac{5}{2})y + (1 + 1)z + (2 + \frac{3}{2}) = 0 \]
This simplifies to:
\[ x - \frac{7}{2}y + 2z + 4 = 0 \implies 2x - 7y + 4z + 7 = 0 \]

### Step 5: Final equations
The equations of the two planes are:
1. \(4x + 7y - 6z - 3 = 0\)
2. \(2x - 7y + 4z + 7 = 0\)

### Step 6: Find the intersection line
The required line is the intersection of these two planes. We can express it in parametric form or symmetric form based on the equations derived.