Find the equations of the line passing through the point (1, -2, 3) and parallel to the planes :
x - y + 2z = 5 and 3x + 2y - z = 6 .

Answer

To find the equations of the line passing through the point (1, -2, 3) and parallel to the given planes, we need to follow these steps:

### Step 1: Identify the normal vectors of the planes
The equations of the planes are given as:
1. \( x - y + 2z = 5 \)
2. \( 3x + 2y - z = 6 \)

The normal vector of a plane \( ax + by + cz = d \) is given by the coefficients \( (a, b, c) \).

- For the first plane \( x - y + 2z = 5 \), the normal vector \( \mathbf{n_1} = (1, -1, 2) \).
- For the second plane \( 3x + 2y - z = 6 \), the normal vector \( \mathbf{n_2} = (3, 2, -1) \).

### Step 2: Find the direction vector of the line
The line we are looking for is parallel to both planes. Thus, its direction vector \( \mathbf{d} \) can be found by taking the cross product of the two normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:

\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 3 & 2 & -1 \end{vmatrix} \]

Calculating this determinant:

\[ \mathbf{d} = \mathbf{i}((-1)(-1) - (2)(2)) - \mathbf{j}((1)(-1) - (2)(3)) + \mathbf{k}((1)(2) - (-1)(3)) \]

\[ = \mathbf{i}(1 - 4) - \mathbf{j}(-1 - 6) + \mathbf{k}(2 + 3) \]

\[ = -3\mathbf{i} + 7\mathbf{j} + 5\mathbf{k} \]

Thus, the direction vector is \( \mathbf{d} = (-3, 7, 5) \).

### Step 3: Write the parametric equations of the line
The line can be expressed in parametric form using the point (1, -2, 3) and the direction vector \( (-3, 7, 5) \).

Let \( t \) be the parameter. The parametric equations are:

\[ x = 1 - 3t \]
\[ y = -2 + 7t \]
\[ z = 3 + 5t \]

### Step 4: Write the symmetric equations of the line
The symmetric equations can be derived from the parametric equations:

\[ \frac{x - 1}{-3} = \frac{y + 2}{7} = \frac{z - 3}{5} \]

### Final Answer
The equations of the line passing through the point (1, -2, 3) and parallel to the planes \( x - y + 2z = 5 \) and \( 3x + 2y - z = 6 \) are:

**Parametric Form:**
\[ x = 1 - 3t, \quad y = -2 + 7t, \quad z = 3 + 5t \]

**Symmetric Form:**
\[ \frac{x - 1}{-3} = \frac{y + 2}{7} = \frac{z - 3}{5} \]