Find the Vector and Cartesian equation of line passing through (1, -2, 3) and parallel to the planes `x - y + 2z = 5 and 3 x + 2y - z = 6`

Answer

To find the vector and Cartesian equations of the line passing through the point \( (1, -2, 3) \) and parallel to the planes defined by the equations \( x - y + 2z = 5 \) and \( 3x + 2y - z = 6 \), we can follow these steps:

### Step 1: Identify the point and the planes
The line passes through the point \( P(1, -2, 3) \). The equations of the planes are:
1. Plane 1: \( x - y + 2z = 5 \)
2. Plane 2: \( 3x + 2y - z = 6 \)

### Step 2: Find the normal vectors of the planes
The normal vector of Plane 1 can be derived from the coefficients of \( x, y, z \) in its equation:
- Normal vector \( \mathbf{n_1} = (1, -1, 2) \)

The normal vector of Plane 2 is:
- Normal vector \( \mathbf{n_2} = (3, 2, -1) \)

### Step 3: Find the direction vector of the line
The line is parallel to both planes, which means its direction vector \( \mathbf{d} \) can be found using the cross product of the normal vectors of the two planes:
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:
\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 3 & 2 & -1 \end{vmatrix} \]

Calculating the 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 4: Write the vector equation of the line
The vector equation of the line can be expressed as:
\[ \mathbf{r} = \mathbf{a} + t\mathbf{d} \]
where \( \mathbf{a} = (1, -2, 3) \) and \( \mathbf{d} = (-3, 7, 5) \).

So, the vector equation is:
\[ \mathbf{r} = (1, -2, 3) + t(-3, 7, 5) \]

### Step 5: Write the Cartesian equation of the line
To convert the vector equation into Cartesian form, we can express it as:
\[ \frac{x - 1}{-3} = \frac{y + 2}{7} = \frac{z - 3}{5} \]

### Final Answers
1. **Vector Equation**:
\[ \mathbf{r} = (1, -2, 3) + t(-3, 7, 5) \]
2. **Cartesian Equation**:
\[ \frac{x - 1}{-3} = \frac{y + 2}{7} = \frac{z - 3}{5} \]