Find the equation of a plane passing through the intersection of the planes `vecr.(2hati-7hatj+4hatk)=3` and `vecr.(3hati-5hatj+4hatk) + 11 - 0` and passes through the point `(-2hati+hatj+3hatk)`.

Answer

To find the equation of the plane passing through the intersection of the given planes and through the specified point, we can follow these steps:

### Step 1: Identify the equations of the given planes
The equations of the two planes are:
1. \( \vec{r} \cdot (2\hat{i} - 7\hat{j} + 4\hat{k}) = 3 \)
2. \( \vec{r} \cdot (3\hat{i} - 5\hat{j} + 4\hat{k}) + 11 = 0 \)

We can rewrite the second equation as:
\[ \vec{r} \cdot (3\hat{i} - 5\hat{j} + 4\hat{k}) = -11 \]

### Step 2: Determine the normal vectors and constants
From the equations, we can identify:
- For the first plane:
- Normal vector \( \vec{n_1} = 2\hat{i} - 7\hat{j} + 4\hat{k} \)
- Constant \( d_1 = 3 \)

- For the second plane:
- Normal vector \( \vec{n_2} = 3\hat{i} - 5\hat{j} + 4\hat{k} \)
- Constant \( d_2 = -11 \)

### Step 3: Write the equation of the plane through the intersection
The equation of the plane passing through the intersection of the two planes can be expressed as:
\[ \vec{r} \cdot \vec{n_1} + \lambda \vec{n_2} = d_1 + \lambda d_2 \]
Substituting the values:
\[ \vec{r} \cdot (2\hat{i} - 7\hat{j} + 4\hat{k}) + \lambda (3\hat{i} - 5\hat{j} + 4\hat{k}) = 3 + \lambda (-11) \]

### Step 4: Simplify the equation
This can be rewritten as:
\[ \vec{r} \cdot \left( (2 + 3\lambda)\hat{i} + (-7 - 5\lambda)\hat{j} + (4 + 4\lambda)\hat{k} \right) = 3 - 11\lambda \]

### Step 5: Substitute the point into the equation
We need to find the value of \( \lambda \) such that the plane passes through the point \( (-2\hat{i} + \hat{j} + 3\hat{k}) \).

Substituting \( \vec{r} = -2\hat{i} + \hat{j} + 3\hat{k} \) into the equation:
\[ (-2)(2 + 3\lambda) + 1(-7 - 5\lambda) + 3(4 + 4\lambda) = 3 - 11\lambda \]

### Step 6: Expand and simplify the equation
Expanding the left side:
\[ -4 - 6\lambda - 7 - 5\lambda + 12 + 12\lambda = 3 - 11\lambda \]
Combining like terms:
\[ (12 - 4 - 7) + (-6\lambda - 5\lambda + 12\lambda) = 3 - 11\lambda \]
This simplifies to:
\[ 1 + \lambda = 3 - 11\lambda \]

### Step 7: Solve for \( \lambda \)
Rearranging gives:
\[ 1 + 11\lambda + \lambda = 3 \]
\[ 12\lambda = 2 \implies \lambda = \frac{1}{6} \]

### Step 8: Substitute \( \lambda \) back into the plane equation
Substituting \( \lambda = \frac{1}{6} \) back into the equation of the plane:
\[ \vec{r} \cdot \left( 2 + 3\left(\frac{1}{6}\right) \right)\hat{i} + \left( -7 - 5\left(\frac{1}{6}\right) \right)\hat{j} + \left( 4 + 4\left(\frac{1}{6}\right) \right)\hat{k} = 3 - 11\left(\frac{1}{6}\right) \]

Calculating each term:
- For \( \hat{i} \): \( 2 + \frac{1}{2} = \frac{5}{2} \)
- For \( \hat{j} \): \( -7 - \frac{5}{6} = -\frac{47}{6} \)
- For \( \hat{k} \): \( 4 + \frac{2}{3} = \frac{14}{3} \)

Thus, the equation becomes:
\[ \vec{r} \cdot \left( \frac{5}{2}\hat{i} - \frac{47}{6}\hat{j} + \frac{14}{3}\hat{k} \right) = \frac{7}{6} \]

### Step 9: Final equation of the plane
To eliminate fractions, multiply through by 6:
\[ \vec{r} \cdot (15\hat{i} - 47\hat{j} + 28\hat{k}) = 7 \]

This is the required equation of the plane.