Find the equation of the plane passing through the point (1, 4, 2) and parallel to the plane 2x + y – 3z – 7 = 0

To find the equation of a plane that is parallel to a given plane and passes through a specific point, we can follow these steps:

1. **Understand the Plane Equation**: The equation of the given plane is in the standard form Ax + By + Cz + D = 0, where the coefficients (A, B, C) represent the normal vector to the plane. For the plane 2x + y – 3z – 7 = 0, the normal vector is (2, 1, -3).

2. **Use the Normal Vector**: Since we need a plane that is parallel to this plane, it must have the same normal vector. Therefore, the equation of the new plane will also have the form 2x + y – 3z + D = 0.

3. **Plug in the Point**: Now we need to find the constant D using the point we’re given, which is (1, 4, 2). We substitute x = 1, y = 4, and z = 2 into the equation:

2(1) + 4 – 3(2) + D = 0

Simplifying this, we have:

2 + 4 – 6 + D = 0

0 + D = 0

So, D = 0.

4. **Write the Final Equation**: Therefore, the equation of the new plane that passes through the point (1, 4, 2) and is parallel to the given plane is:

2x + y – 3z = 0.

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