Find the equation of plane passing through the line of intersection of planes `vecr.(hati+3hatj) +6=0 and vecr.(3hati-hatj-4hatk)=0` , whose perpendicular distance from origin is one unit.

Answer

To find the equation of the plane passing through the line of intersection of the given planes and having a perpendicular distance of one unit from the origin, we can follow these steps:

### Step 1: Identify the equations of the given planes
The equations of the planes are:
1. \( \vec{r} \cdot (\hat{i} + 3\hat{j}) + 6 = 0 \)
2. \( \vec{r} \cdot (3\hat{i} - \hat{j} - 4\hat{k}) = 0 \)

### Step 2: Write the family of planes passing through the line of intersection
The equation of the family of planes passing through the line of intersection of the two planes can be expressed as:
\[ \vec{r} \cdot (\hat{i} + 3\hat{j}) + 6 + \lambda (\vec{r} \cdot (3\hat{i} - \hat{j} - 4\hat{k})) = 0 \]
This simplifies to:
\[ \vec{r} \cdot (\hat{i} + 3\hat{j} + \lambda(3\hat{i} - \hat{j} - 4\hat{k})) = -6 \]

### Step 3: Combine the terms
Combining the terms inside the dot product:
\[ \vec{r} \cdot ((1 + 3\lambda)\hat{i} + (3 - \lambda)\hat{j} - 4\lambda\hat{k}) = -6 \]

### Step 4: Set up the equation for the perpendicular distance
The perpendicular distance \(d\) from the origin to the plane given by \(Ax + By + Cz + D = 0\) is given by:
\[ d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \]
In our case, \(D = -6\), \(A = 1 + 3\lambda\), \(B = 3 - \lambda\), and \(C = -4\lambda\). We need the distance to be 1 unit:
\[ 1 = \frac{6}{\sqrt{(1 + 3\lambda)^2 + (3 - \lambda)^2 + (-4\lambda)^2}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 1 = \frac{36}{(1 + 3\lambda)^2 + (3 - \lambda)^2 + (-4\lambda)^2} \]
Rearranging this gives:
\[ (1 + 3\lambda)^2 + (3 - \lambda)^2 + (-4\lambda)^2 = 36 \]

### Step 6: Expand and simplify the equation
Expanding each term:
\[ (1 + 3\lambda)^2 = 1 + 6\lambda + 9\lambda^2 \]
\[ (3 - \lambda)^2 = 9 - 6\lambda + \lambda^2 \]
\[ (-4\lambda)^2 = 16\lambda^2 \]
Combining these:
\[ 1 + 6\lambda + 9\lambda^2 + 9 - 6\lambda + \lambda^2 + 16\lambda^2 = 36 \]
This simplifies to:
\[ 26\lambda^2 + 10 = 36 \]
Thus:
\[ 26\lambda^2 = 26 \implies \lambda^2 = 1 \implies \lambda = \pm 1 \]

### Step 7: Find the equations of the planes for both values of \(\lambda\)
1. For \(\lambda = 1\):
\[ \vec{r} \cdot (1 + 3 \cdot 1)\hat{i} + (3 - 1)\hat{j} - 4\cdot 1\hat{k} = -6 \]
This gives:
\[ \vec{r} \cdot (4\hat{i} + 2\hat{j} - 4\hat{k}) = -6 \]
Or:
\[ 4x + 2y - 4z = -6 \]

2. For \(\lambda = -1\):
\[ \vec{r} \cdot (1 - 3)\hat{i} + (3 + 1)\hat{j} + 4\hat{k} = -6 \]
This gives:
\[ \vec{r} \cdot (-2\hat{i} + 4\hat{j} + 4\hat{k}) = -6 \]
Or:
\[ -2x + 4y + 4z = -6 \]

### Final Answer
The equations of the planes are:
1. \(4x + 2y - 4z = -6\)
2. \(-2x + 4y + 4z = -6\)