The equation of line which passes through the intersection of the planes `x+2y-3z-4=0 and 3x-8y+z+2=0` is

Options

A. `(x-2)/22=(y-1)/10=(z-0)/14`
B. `(x-2)/-22=(y-1)/-10=(z-0)/-14`
C. `(x+2)/22=(y+1)/10=(z-0)/14`
D. None of these

Answer

To find the equation of the line that passes through the intersection of the two planes given by the equations \(x + 2y - 3z - 4 = 0\) and \(3x - 8y + z + 2 = 0\), we can follow these steps:

### Step 1: Write the equations of the planes in vector form
The equations of the planes can be written in vector form as follows:
1. For the first plane:
\[ \mathbf{n_1} = \langle 1, 2, -3 \rangle \]
This corresponds to the equation \(x + 2y - 3z - 4 = 0\).

2. For the second plane:
\[ \mathbf{n_2} = \langle 3, -8, 1 \rangle \]
This corresponds to the equation \(3x - 8y + z + 2 = 0\).

### Step 2: Find the direction vector of the line
To find the direction vector of the line that is the intersection of the two planes, we take the cross product of the normal vectors of the planes:
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]
Calculating the cross product:
\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 3 & -8 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{d} = \mathbf{i} \begin{vmatrix} 2 & -3 \\ -8 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -3 \\ 3 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 3 & -8 \end{vmatrix} \]
Calculating each of the determinants:
- For \( \mathbf{i} \): \(2 \cdot 1 - (-3)(-8) = 2 - 24 = -22\)
- For \( \mathbf{j} \): \(1 \cdot 1 - (-3)(3) = 1 + 9 = 10\)
- For \( \mathbf{k} \): \(1 \cdot (-8) - 2 \cdot 3 = -8 - 6 = -14\)

Thus,
\[ \mathbf{d} = \langle -22, -10, -14 \rangle \]

### Step 3: Find a point on the line
To find a point on the line, we can set \(z = 0\) in the equations of the planes and solve for \(x\) and \(y\).

Substituting \(z = 0\) in the first plane:
\[ x + 2y - 4 = 0 \implies x + 2y = 4 \quad \text{(1)} \]

Substituting \(z = 0\) in the second plane:
\[ 3x - 8y + 2 = 0 \implies 3x - 8y = -2 \quad \text{(2)} \]

Now we can solve equations (1) and (2) simultaneously.

From equation (1):
\[ x = 4 - 2y \]

Substituting into equation (2):
\[ 3(4 - 2y) - 8y = -2 \]
\[ 12 - 6y - 8y = -2 \]
\[ 12 - 14y = -2 \]
\[ -14y = -14 \implies y = 1 \]

Substituting \(y = 1\) back into equation (1):
\[ x + 2(1) = 4 \implies x + 2 = 4 \implies x = 2 \]

Thus, the point on the line is \((2, 1, 0)\).

### Step 4: Write the equation of the line
The equation of the line in symmetric form is given by:
\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \]
Where \((x_0, y_0, z_0)\) is a point on the line and \((a, b, c)\) is the direction vector.

Using the point \((2, 1, 0)\) and direction vector \((-22, -10, -14)\):
\[ \frac{x - 2}{-22} = \frac{y - 1}{-10} = \frac{z - 0}{-14} \]

### Final Answer
The equation of the line is:
\[ \frac{x - 2}{-22} = \frac{y - 1}{-10} = \frac{z}{-14} \]