Find the equation of the plane through the point (1,4,-2) and parallel to the plane `-2x+y-3z=7`.

Answer

To find the equation of the plane through the point (1, 4, -2) and parallel to the plane given by the equation \(-2x + y - 3z = 7\), we can follow these steps:

### Step 1: Identify the normal vector of the given plane
The equation of the plane can be rewritten in the standard form \(Ax + By + Cz + D = 0\). For the given plane \(-2x + y - 3z = 7\), we can rearrange it to:
\[ -2x + y - 3z - 7 = 0 \]
From this, we can identify the coefficients \(A = -2\), \(B = 1\), and \(C = -3\). The normal vector \(\vec{n}\) of the plane is given by:
\[ \vec{n} = (-2, 1, -3) \]

### Step 2: Write the general equation of a parallel plane
Since the plane we want to find is parallel to the given plane, it will have the same normal vector. Thus, the equation of the plane can be expressed as:
\[ -2x + y - 3z + k = 0 \]
where \(k\) is a constant that we need to determine.

### Step 3: Substitute the point into the plane equation
We know that the plane must pass through the point (1, 4, -2). We will substitute these coordinates into the plane equation to find \(k\):
\[ -2(1) + 1(4) - 3(-2) + k = 0 \]
Calculating this gives:
\[ -2 + 4 + 6 + k = 0 \]
\[ 8 + k = 0 \]

### Step 4: Solve for \(k\)
Now, we can solve for \(k\):
\[ k = -8 \]

### Step 5: Write the final equation of the plane
Substituting \(k\) back into the equation of the plane, we get:
\[ -2x + y - 3z - 8 = 0 \]
To express it in a more standard form, we can rearrange it:
\[ -2x + y - 3z = 8 \]
or multiplying through by -1:
\[ 2x - y + 3z = -8 \]

Thus, the equation of the plane is:
\[ 2x - y + 3z = -8 \]