Find the equation of the plane passes through the point `(1,-3,1)` and parallel to the plane `2x+3y+z=1`.

Answer

To find the equation of the plane that passes through the point \( (1, -3, 1) \) and is parallel to the plane given by the equation \( 2x + 3y + z = 1 \), we can follow these steps:

### Step 1: Identify the normal vector of the given plane
The equation of the plane \( 2x + 3y + z = 1 \) can be rewritten in the standard form \( Ax + By + Cz + D = 0 \). Here, we can identify:
- \( A = 2 \)
- \( B = 3 \)
- \( C = 1 \)

The normal vector \( \vec{n} \) of the plane is given by the coefficients of \( x, y, z \):
\[ \vec{n} = (2, 3, 1) \]

### Step 2: Use the point-normal form of the plane equation
Since the required plane is parallel to the given plane, it will have the same normal vector. The point-normal form of the equation of a plane is given by:
\[ A(x - x_1) + B(y - y_1) + C(z - z_1) = 0 \]
where \( (x_1, y_1, z_1) \) is a point on the plane.

### Step 3: Substitute the values into the equation
We know the point \( (1, -3, 1) \) lies on the plane, so we substitute \( x_1 = 1 \), \( y_1 = -3 \), and \( z_1 = 1 \) into the equation:
\[ 2(x - 1) + 3(y + 3) + 1(z - 1) = 0 \]

### Step 4: Expand the equation
Now, we expand the equation:
\[ 2x - 2 + 3y + 9 + z - 1 = 0 \]

### Step 5: Combine like terms
Combining the constant terms:
\[ 2x + 3y + z + 6 = 0 \]

### Step 6: Rearrange to standard form
We can rearrange this to the standard form of the plane equation:
\[ 2x + 3y + z = -6 \]

Thus, the equation of the plane that passes through the point \( (1, -3, 1) \) and is parallel to the plane \( 2x + 3y + z = 1 \) is:
\[ 2x + 3y + z = -6 \]