Find the equation of the plane through the intersection of the planes `x+3y+6=0 and 3x-y-4z=0`, whose perpendicular distance from the origin is unity.

Answer

To find the equation of the plane through the intersection of the planes \(x + 3y + 6 = 0\) and \(3x - y - 4z = 0\), whose perpendicular distance from the origin is unity, we can follow these steps:

### Step 1: Write the equation of the plane through the intersection of the given planes.

The equation of the plane through the intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
where \(P_1\) and \(P_2\) are the equations of the given planes. Thus, we have:
\[ (x + 3y + 6) + \lambda(3x - y - 4z) = 0 \]
Expanding this gives:
\[ x + 3y + 6 + \lambda(3x - y - 4z) = 0 \]
This simplifies to:
\[ (1 + 3\lambda)x + (3 - \lambda)y + (0 - 4\lambda)z + 6 = 0 \]

### Step 2: Identify coefficients.

From the equation, we can identify:
- Coefficient of \(x\): \(A = 1 + 3\lambda\)
- Coefficient of \(y\): \(B = 3 - \lambda\)
- Coefficient of \(z\): \(C = -4\lambda\)
- Constant term: \(D = 6\)

### Step 3: Use the formula for the perpendicular distance from the origin.

The perpendicular distance \(d\) from the origin to the plane \(Ax + By + Cz + D = 0\) is given by:
\[ d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \]
Given that \(d = 1\), we can set up the equation:
\[ \frac{|6|}{\sqrt{(1 + 3\lambda)^2 + (3 - \lambda)^2 + (-4\lambda)^2}} = 1 \]

### Step 4: Square both sides to eliminate the square root.

Squaring both sides gives:
\[ 36 = (1 + 3\lambda)^2 + (3 - \lambda)^2 + (4\lambda)^2 \]

### Step 5: Expand and simplify.

Expanding the squares:
\[ (1 + 3\lambda)^2 = 1 + 6\lambda + 9\lambda^2 \]
\[ (3 - \lambda)^2 = 9 - 6\lambda + \lambda^2 \]
\[ (-4\lambda)^2 = 16\lambda^2 \]
Combining these:
\[ 36 = (1 + 6\lambda + 9\lambda^2) + (9 - 6\lambda + \lambda^2) + (16\lambda^2) \]
This simplifies to:
\[ 36 = 10 + 26\lambda^2 \]
So we have:
\[ 26\lambda^2 = 26 \]
Thus:
\[ \lambda^2 = 1 \implies \lambda = \pm 1 \]

### Step 6: Substitute \(\lambda\) back into the plane equation.

1. For \(\lambda = 1\):
\[ (1 + 3(1))x + (3 - 1)y - 4(1)z + 6 = 0 \]
This simplifies to:
\[ 4x + 2y - 4z + 6 = 0 \implies 2x + y - 2z + 3 = 0 \]

2. For \(\lambda = -1\):
\[ (1 + 3(-1))x + (3 - (-1))y - 4(-1)z + 6 = 0 \]
This simplifies to:
\[ -2x + 4y + 4z + 6 = 0 \implies x - 2y - 2z - 3 = 0 \]

### Final Answer:
The equations of the required planes are:
1. \(2x + y - 2z + 3 = 0\)
2. \(x - 2y - 2z - 3 = 0\)