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To solve this problem, we need to find the equation of a plane passing through the line of intersection of two given planes and having a perpendicular distance from the origin equal to 1 (unity).

### Given:
1. Two planes:
   \[
   \text{Plane 1: } \mathbf{r} \cdot (\mathbf{i} + 3\mathbf{j}) - 6 = 0
   \]
   \[
   \text{Plane 2: } \mathbf{r} \cdot (3\mathbf{i} - \mathbf{j} - 4\mathbf{k}) = 0
   \]

2. Required: Equation of a plane passing through their line of intersection with a perpendicular distance from the origin equal to 1.

### Step-by-Step Solution:

#### Step 1: General Equation of the Plane Passing Through the Line of Intersection
The equation of a plane passing through the line of intersection of the given planes can be written as:

\[
\mathbf{r} \cdot (\mathbf{i} + 3\mathbf{j}) - 6 + \lambda [\mathbf{r} \cdot (3\mathbf{i} - \mathbf{j} - 4\mathbf{k})] = 0
\]

Simplify the expression:

\[
(\mathbf{r} \cdot \mathbf{i} + 3\mathbf{r} \cdot \mathbf{j}) - 6 + \lambda (3\mathbf{r} \cdot \mathbf{i} - \mathbf{r} \cdot \mathbf{j} - 4\mathbf{r} \cdot \mathbf{k}) = 0
\]

Substitute \( \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \):

\[
(x + 3y - 6) + \lambda (3x - y - 4z) = 0
\]

Combine like terms:

\[
(x + 3\lambda x) + (3y - \lambda y) + (-4\lambda z) - 6 = 0
\]

This simplifies to:

\[
(1 + 3\lambda)x + (3 - \lambda)y - 4\lambda z - 6 = 0
\]

#### Step 2: Find the Value of \( \lambda \) Such That Perpendicular Distance from the Origin is Unity

The perpendicular distance of the plane \( Ax + By + Cz + D = 0 \) from the origin is given by:

\[
\frac{|D|}{\sqrt{A^2 + B^2 + C^2}} = 1
\]

For our equation, \( A = 1 + 3\lambda \), \( B = 3 - \lambda \), \( C = -4\lambda \), and \( D = -6 \). Thus:

\[
\frac{| -6 |}{\sqrt{(1 + 3\lambda)^2 + (3 - \lambda)^2 + (-4\lambda)^2}} = 1
\]

Simplify:

\[
\frac{6}{\sqrt{(1 + 3\lambda)^2 + (3 - \lambda)^2 + 16\lambda^2}} = 1
\]

Square both sides:

\[
36 = (1 + 3\lambda)^2 + (3 - \lambda)^2 + 16\lambda^2
\]

Expand and simplify:

\[
36 = (1 + 6\lambda + 9\lambda^2) + (9 - 6\lambda + \lambda^2) + 16\lambda^2
\]

Combine like terms:

\[
36 = 1 + 9 + 6\lambda - 6\lambda + 9\lambda^2 + \lambda^2 + 16\lambda^2
\]

\[
36 = 10 + 26\lambda^2
\]

Solve for \( \lambda^2 \):

\[
26\lambda^2 = 26 \implies \lambda^2 = 1 \implies \lambda = \pm 1
\]

#### Step 3: Substitute the Values of \( \lambda \) to Find the Plane Equations

1. For \( \lambda = 1 \):

\[
(1 + 3 \times 1)x + (3 - 1)y - 4 \times 1z - 6 = 0 \implies 4x + 2y

Find the equation of the plane passing through the line of intersection of the planes r ⋅ (𝑖 + 3𝑗) - 6 = 0 and r ⋅ (3𝑖 − 𝑗 − 4𝑘) = 0 whose perpendicular distance from the origin is unity.

Learn how to find the equation of a plane passing through the line of intersection of two given planes, ensuring the perpendicular distance from the origin is unity. This detailed solution explains key concepts like vector algebra, planes, and distance formulas.

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