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

Answer

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

### Step 1: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \( Ax + By + Cz = D \) is \( (A, B, C) \).

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

### Step 2: Find the direction ratios of the line
Since the line is parallel to both planes, its direction ratios \( (a, b, c) \) must be perpendicular to the normal vectors of both planes. This gives us two equations:

1. From the first plane:
\[ a - b + 2c = 0 \]

2. From the second plane:
\[ 3a + b + c = 0 \]

### Step 3: Solve the system of equations
We now have a system of linear equations:
1. \( a - b + 2c = 0 \) (Equation 1)
2. \( 3a + b + c = 0 \) (Equation 2)

We can solve these equations simultaneously.

From Equation 1, we can express \( b \) in terms of \( a \) and \( c \):
\[ b = a + 2c \]

Substituting \( b \) into Equation 2:
\[ 3a + (a + 2c) + c = 0 \]
\[ 4a + 3c = 0 \]
From this, we can express \( c \) in terms of \( a \):
\[ c = -\frac{4}{3}a \]

Now substituting \( c \) back into the expression for \( b \):
\[ b = a + 2\left(-\frac{4}{3}a\right) = a - \frac{8}{3}a = -\frac{5}{3}a \]

### Step 4: Write the direction ratios
We can choose \( a = 3 \) (to avoid fractions), then:
\[ b = -5, \quad c = -4 \]
Thus, the direction ratios of the line are \( (3, -5, -4) \).

### Step 5: Write the equation of the line
The equation of the line passing through the point \( (1, 2, 3) \) with direction ratios \( (3, -5, -4) \) is given by:
\[ \frac{x - 1}{3} = \frac{y - 2}{-5} = \frac{z - 3}{-4} \]

### Final Answer
The equation of the line is:
\[ \frac{x - 1}{3} = \frac{y - 2}{-5} = \frac{z - 3}{-4} \]