Find the equation of the plane passes through the point `(2,3,5)` and parallel to the plane `x-3y+z=8`.

Answer

To find the equation of the plane that passes through the point \( (2, 3, 5) \) and is parallel to the plane given by the equation \( x - 3y + z = 8 \), we can follow these steps:

### Step 1: Identify the normal vector of the given plane
The equation of the plane \( x - 3y + z = 8 \) can be rewritten in the form \( Ax + By + Cz + D = 0 \). Here, we can identify the coefficients:
- \( A = 1 \)
- \( B = -3 \)
- \( C = 1 \)

The normal vector \( \mathbf{n} \) of the given plane is given by:
\[ \mathbf{n} = (A, B, C) = (1, -3, 1) \]

### Step 2: Use the point-normal form of the plane equation
The point-normal form of the equation of a plane is given by:
\[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{a}) = 0 \]
where \( \mathbf{r} = (x, y, z) \) is a position vector of any point on the plane, and \( \mathbf{a} = (x_0, y_0, z_0) \) is a position vector of a known point on the plane.

Given the point \( (2, 3, 5) \), we have:
\[ \mathbf{a} = (2, 3, 5) \]

### Step 3: Substitute the normal vector and point into the equation
Substituting \( \mathbf{n} = (1, -3, 1) \) and \( \mathbf{a} = (2, 3, 5) \) into the point-normal form:
\[ (1, -3, 1) \cdot ((x, y, z) - (2, 3, 5)) = 0 \]

This expands to:
\[ (1)(x - 2) + (-3)(y - 3) + (1)(z - 5) = 0 \]

### Step 4: Simplify the equation
Expanding the equation:
\[ x - 2 - 3y + 9 + z - 5 = 0 \]
Combine like terms:
\[ x - 3y + z + 2 = 0 \]

### Step 5: Rearrange to standard form
Rearranging gives us the equation of the plane:
\[ x - 3y + z = -2 \]

### Final Answer
The equation of the plane that passes through the point \( (2, 3, 5) \) and is parallel to the plane \( x - 3y + z = 8 \) is:
\[ x - 3y + z = -2 \]