What are the direction ratios of the line of intersection of the planes `x=3z+4 and y=2z-3`?

Options

A. `(:1,2,3:)`
B. `(:2,1,3:)`
C. `(:3,2,1:)`
D. `(:1,3,2:)`

Explanation

To find the direction ratios of the line of intersection of the planes given by the equations \( x = 3z + 4 \) and \( y = 2z - 3 \), we can follow these steps:

### Step 1: Rewrite the equations of the planes
The equations of the planes can be rewritten in standard form:
1. For the plane \( x = 3z + 4 \), we can rearrange it to:
\[ x - 3z - 4 = 0 \]
This gives us the normal vector \( \mathbf{n_1} = (1, 0, -3) \).

2. For the plane \( y = 2z - 3 \), we can rearrange it to:
\[ y - 2z + 3 = 0 \]
This gives us the normal vector \( \mathbf{n_2} = (0, 1, -2) \).

### Step 2: Find the direction ratios of the line of intersection
The direction ratios of the line of intersection of two planes can be found using the cross product of their normal vectors.

Let:
\[ \mathbf{n_1} = (1, 0, -3) \]
\[ \mathbf{n_2} = (0, 1, -2) \]

The cross product \( \mathbf{n_1} \times \mathbf{n_2} \) is computed as follows:
\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -3 \\ 0 & 1 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we get:
\[ \mathbf{n_1} \times \mathbf{n_2} = \mathbf{i} \begin{vmatrix} 0 & -3 \\ 1 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -3 \\ 0 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. For \( \mathbf{i} \):
\[ 0 \cdot (-2) - (-3) \cdot 1 = 3 \]
2. For \( \mathbf{j} \):
\[ 1 \cdot (-2) - (-3) \cdot 0 = -2 \]
3. For \( \mathbf{k} \):
\[ 1 \cdot 1 - 0 \cdot 0 = 1 \]

Putting it all together, we have:
\[ \mathbf{n_1} \times \mathbf{n_2} = (3, 2, 1) \]

### Step 4: Conclusion
Thus, the direction ratios of the line of intersection of the given planes are \( (3, 2, 1) \).