Find the equation of the plane which is perpendicualr to the plane `vecr(5hati+2hatj+6hatk)+8=0` and which is containg the line of intersection of the planes `vecr(hati+2hatj+3hatk)=4` and `vecr(2hati+hatj-hatk)+5=0`

Answer

To find the equation of the plane that is perpendicular to the given plane and contains the line of intersection of two other planes, we can follow these steps:

### Step 1: Identify the normal vector of the given plane
The equation of the given plane is:
\[ \vec{r} \cdot (5\hat{i} + 2\hat{j} + 6\hat{k}) + 8 = 0 \]
From this, we can identify the normal vector \( \vec{n_1} = 5\hat{i} + 2\hat{j} + 6\hat{k} \).

### Step 2: Write the equations of the two planes
The two planes given are:
1. \( \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 4 \) which can be rewritten as:
\[ \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) - 4 = 0 \]
Here, the normal vector is \( \vec{n_2} = \hat{i} + 2\hat{j} + 3\hat{k} \).

2. \( \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) + 5 = 0 \) which can be rewritten as:
\[ \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) + 5 = 0 \]
Here, the normal vector is \( \vec{n_3} = 2\hat{i} + \hat{j} - \hat{k} \).

### Step 3: Find the equation of the plane containing the line of intersection
The equation of the required plane \( P_3 \) can be expressed as:
\[ P_3 = P_1 + \lambda P_2 \]
Substituting the equations of the planes:
\[ \vec{r} \cdot (5\hat{i} + 2\hat{j} + 6\hat{k}) - 4 + \lambda \left( \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) + 5 \right) = 0 \]

### Step 4: Combine the equations
This gives us:
\[ \vec{r} \cdot (5\hat{i} + 2\hat{j} + 6\hat{k} + \lambda (2\hat{i} + \hat{j} - \hat{k})) - (4 + 5\lambda) = 0 \]
Let’s denote the new normal vector as:
\[ \vec{n_4} = (5 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (6 - \lambda)\hat{k} \]

### Step 5: Set the condition for perpendicularity
For the required plane to be perpendicular to the given plane, the dot product of their normal vectors must be zero:
\[ \vec{n_1} \cdot \vec{n_4} = 0 \]
Calculating the dot product:
\[ (5)(5 + 2\lambda) + (2)(2 + \lambda) + (6)(6 - \lambda) = 0 \]

### Step 6: Solve for \( \lambda \)
Expanding this:
\[ 25 + 10\lambda + 4 + 2\lambda + 36 - 6\lambda = 0 \]
Combining like terms:
\[ (10\lambda + 2\lambda - 6\lambda) + (25 + 4 + 36) = 0 \]
This simplifies to:
\[ 6\lambda + 65 = 0 \]
Thus,
\[ 6\lambda = -65 \implies \lambda = -\frac{65}{6} \]

### Step 7: Substitute \( \lambda \) back to find the equation of the plane
Substituting \( \lambda \) back into the normal vector:
\[ \vec{n_4} = \left( 5 + 2\left(-\frac{65}{6}\right) \right)\hat{i} + \left( 2 - \frac{65}{6} \right)\hat{j} + \left( 6 + \frac{65}{6} \right)\hat{k} \]
Calculating each component:
1. \( 5 + 2\left(-\frac{65}{6}\right) = 5 - \frac{130}{6} = \frac{30 - 130}{6} = -\frac{100}{6} = -\frac{50}{3} \)
2. \( 2 - \frac{65}{6} = \frac{12 - 65}{6} = -\frac{53}{6} \)
3. \( 6 + \frac{65}{6} = \frac{36 + 65}{6} = \frac{101}{6} \)

Thus, the normal vector becomes:
\[ \vec{n_4} = -\frac{50}{3}\hat{i} - \frac{53}{6}\hat{j} + \frac{101}{6}\hat{k} \]

### Step 8: Write the final equation of the plane
The equation of the plane can be expressed as:
\[ \vec{r} \cdot \left(-\frac{50}{3}\hat{i} - \frac{53}{6}\hat{j} + \frac{101}{6}\hat{k}\right) = d \]
Where \( d \) is calculated from substituting a point on the line of intersection (which can be derived from the two planes).