Find the equation of planes parallel to the plane `2x-4y+4z=7` and which are at a distance of five units from the point `(3,-1,2)`

Answer

To find the equation of planes parallel to the plane \(2x - 4y + 4z = 7\) and which are at a distance of five units from the point \((3, -1, 2)\), we can follow these steps:

### Step 1: Identify the Normal Vector
The normal vector of the given plane \(2x - 4y + 4z = 7\) is \(\vec{n} = (2, -4, 4)\). Since we are looking for planes parallel to this one, the new planes will have the same normal vector.

### Step 2: Write the General Form of the Plane
The equation of a plane parallel to the given plane can be expressed in the form:
\[ 2x - 4y + 4z = d \]
where \(d\) is a constant that we need to determine.

### Step 3: Use the Distance Formula
The distance \(D\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz = D\) is given by the formula:
\[ D = \frac{|Ax_1 + By_1 + Cz_1 - d|}{\sqrt{A^2 + B^2 + C^2}} \]
In our case, \(A = 2\), \(B = -4\), \(C = 4\), and the point is \((3, -1, 2)\). We want this distance to equal 5 units.

### Step 4: Substitute the Values into the Distance Formula
Substituting the values into the distance formula, we have:
\[ 5 = \frac{|2(3) - 4(-1) + 4(2) - d|}{\sqrt{2^2 + (-4)^2 + 4^2}} \]
Calculating the denominator:
\[ \sqrt{2^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \]
Thus, we can rewrite the equation as:
\[ 5 = \frac{|6 + 4 + 8 - d|}{6} \]
This simplifies to:
\[ 5 = \frac{|18 - d|}{6} \]

### Step 5: Solve for \(d\)
Multiplying both sides by 6 gives:
\[ 30 = |18 - d| \]
This absolute value equation leads to two cases:
1. \(18 - d = 30\)
2. \(18 - d = -30\)

#### Case 1:
\[ 18 - d = 30 \implies d = 18 - 30 = -12 \]

#### Case 2:
\[ 18 - d = -30 \implies d = 18 + 30 = 48 \]

### Step 6: Write the Equations of the Planes
Now we have two values for \(d\):
1. For \(d = -12\), the equation of the plane is:
\[ 2x - 4y + 4z = -12 \]
2. For \(d = 48\), the equation of the plane is:
\[ 2x - 4y + 4z = 48 \]

### Final Answer
The equations of the planes are:
1. \(2x - 4y + 4z + 12 = 0\)
2. \(2x - 4y + 4z - 48 = 0\)