Find the equation of the plane , which contains the line of intersection of the planes :
`vec(r). (hati + 2 hatj + 3 hatk) -4 = 0 and vec(r). (2 hati + hatj + hatk) + 5 = 0 ` and which is perpendicular to the plane :
`vec(r) . (5 hati + 3 hatj - 6 hatk) ) + 8 = 0` .

Answer

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

### Step 1: Identify the equations of the given planes

The equations of the two planes are:
1. \( P_1: \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) - 4 = 0 \)
2. \( P_2: \vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) + 5 = 0 \)

### Step 2: Write the normal vectors of the planes

From the equations, we can extract the normal vectors of the planes:
- For \( P_1 \), the normal vector \( \vec{n_1} = \hat{i} + 2\hat{j} + 3\hat{k} \)
- For \( P_2 \), the normal vector \( \vec{n_2} = 2\hat{i} + \hat{j} + \hat{k} \)

### Step 3: Find the equation of the family of planes containing the line of intersection

The equation of the family of planes containing the line of intersection of \( P_1 \) and \( P_2 \) can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations:
\[ (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + \hat{j} + \hat{k}) = 0 \]
This leads to:
\[ (1 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 + \lambda)\hat{k} = 0 \]

### Step 4: Determine the normal vector of the family of planes

The normal vector of the family of planes is:
\[ \vec{n} = (1 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 + \lambda)\hat{k} \]

### Step 5: Find the equation of the plane perpendicular to the given plane

The plane we are looking for is perpendicular to the plane:
\[ P_3: \vec{r} \cdot (5\hat{i} + 3\hat{j} - 6\hat{k}) + 8 = 0 \]
Thus, its normal vector is \( \vec{n_3} = 5\hat{i} + 3\hat{j} - 6\hat{k} \).

For the planes to be perpendicular, the dot product of their normal vectors must be zero:
\[ \vec{n} \cdot \vec{n_3} = 0 \]
Substituting the normal vector:
\[ (1 + 2\lambda) \cdot 5 + (2 + \lambda) \cdot 3 + (3 + \lambda) \cdot (-6) = 0 \]

### Step 6: Solve for \( \lambda \)

Expanding this gives:
\[ 5 + 10\lambda + 6 + 3\lambda - 18 - 6\lambda = 0 \]
Combining like terms:
\[ (10\lambda + 3\lambda - 6\lambda) + (5 + 6 - 18) = 0 \]
This simplifies to:
\[ 7\lambda - 7 = 0 \]
Thus, we find:
\[ \lambda = 1 \]

### Step 7: Substitute \( \lambda \) back into the normal vector

Substituting \( \lambda = 1 \) into the normal vector:
\[ \vec{n} = (1 + 2 \cdot 1)\hat{i} + (2 + 1)\hat{j} + (3 + 1)\hat{k} = 3\hat{i} + 3\hat{j} + 4\hat{k} \]

### Step 8: Write the equation of the plane

The equation of the plane can be written as:
\[ 3x + 3y + 4z + D = 0 \]
To find \( D \), we can substitute a point on the line of intersection of the two original planes or use the constant from the family of planes. Using the constant from the family of planes when \( \lambda = 1 \):
\[ D = 1 - 4 = -3 \]
Thus, the equation of the plane is:
\[ 3x + 3y + 4z + 1 = 0 \]

### Final Answer

The equation of the required plane is:
\[ 3x + 3y + 4z + 1 = 0 \]