Find the equation of a plane through the intersection of the planes :
`vec(r) ( 2 hati + hatj + 3 hatk) = 7 and vec(r). (2 hati + 3 hatj + 3 hatk) = 9 `
and passing through the point (2,1,3).

Answer

To find the equation of the plane through the intersection of the given planes and passing through the point (2, 1, 3), we can follow these steps:

### Step 1: Write the equations of the given planes in standard form.

The equations of the planes are given as:
1. \( \vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) = 7 \)
2. \( \vec{r} \cdot (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) = 9 \)

We can rewrite these equations as:
1. \( \vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) - 7 = 0 \)
2. \( \vec{r} \cdot (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - 9 = 0 \)

Let:
- \( P_1 = \vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) - 7 \)
- \( P_2 = \vec{r} \cdot (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - 9 \)

### Step 2: Form the equation of the required plane.

The equation of the required plane can be expressed as a linear combination of the two planes:
\[ P = P_1 + \lambda P_2 = 0 \]
Substituting for \( P_1 \) and \( P_2 \):
\[ \left( \vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) - 7 \right) + \lambda \left( \vec{r} \cdot (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - 9 \right) = 0 \]

### Step 3: Expand the equation.

Expanding the equation gives:
\[ \vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) + \lambda \vec{r} \cdot (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - 7 - 9\lambda = 0 \]

### Step 4: Combine like terms.

Combining the terms yields:
\[ \vec{r} \cdot \left( (2 + 2\lambda) \hat{i} + (1 + 3\lambda) \hat{j} + (3 + 3\lambda) \hat{k} \right) - (7 + 9\lambda) = 0 \]

### Step 5: Substitute the point (2, 1, 3).

We know the plane passes through the point (2, 1, 3). We can substitute \( \vec{r} = 2\hat{i} + 1\hat{j} + 3\hat{k} \) into the equation:
\[ (2 + 2\lambda)(2) + (1 + 3\lambda)(1) + (3 + 3\lambda)(3) - (7 + 9\lambda) = 0 \]

### Step 6: Simplify the equation.

Calculating the left-hand side:
\[ (2 + 2\lambda)(2) = 4 + 4\lambda \]
\[ (1 + 3\lambda)(1) = 1 + 3\lambda \]
\[ (3 + 3\lambda)(3) = 9 + 9\lambda \]
Combining these:
\[ 4 + 4\lambda + 1 + 3\lambda + 9 + 9\lambda - 7 - 9\lambda = 0 \]
This simplifies to:
\[ (4\lambda + 3\lambda + 9\lambda - 9\lambda) + (4 + 1 + 9 - 7) = 0 \]
\[ (7\lambda) + 7 = 0 \]
Thus:
\[ 7\lambda + 7 = 0 \implies \lambda = -1 \]

### Step 7: Substitute \( \lambda \) back into the plane equation.

Substituting \( \lambda = -1 \) back into the equation:
\[ \vec{r} \cdot \left( (2 - 2) \hat{i} + (1 - 3) \hat{j} + (3 - 3) \hat{k} \right) - (7 - 9) = 0 \]
This simplifies to:
\[ \vec{r} \cdot (0 \hat{i} - 2 \hat{j} + 0 \hat{k}) + 2 = 0 \]
Thus:
\[ -2y + 2 = 0 \implies y = 1 \]

### Final Equation of the Plane:

The equation of the plane is:
\[ y + 1 = 0 \]