Find the equation of the plane through the intersection of the planes. `vecr. (hati+3hatj-hatk)=9 and vecr. (2hati-hatj+hatk)=3 and` passing through the origin.

Answer

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

### Step 1: Identify the normal vectors and constants from the given plane equations.

The equations of the planes are:

1. \(\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 9\)
2. \(\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3\)

From these equations, we can identify the normal vectors and the constants:

- For the first plane, the normal vector \(\vec{n_1} = \hat{i} + 3\hat{j} - \hat{k}\) and \(d_1 = 9\).
- For the second plane, the normal vector \(\vec{n_2} = 2\hat{i} - \hat{j} + \hat{k}\) and \(d_2 = 3\).

### Step 2: Write the general equation of the plane through the intersection of the two planes.

The equation of the plane through the intersection of the two planes can be expressed as:

\[ \vec{r} \cdot \vec{n_1} + \lambda \vec{r} \cdot \vec{n_2 = d_1 + \lambda d_2} \]

Substituting the values we have:

\[ \vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) + \lambda \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 9 + 3\lambda \]

### Step 3: Substitute \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).

Substituting \(\vec{r}\) into the equation gives:

\[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 3\hat{j} - \hat{k}) + \lambda (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k}) = 9 + 3\lambda \]

Calculating the dot products:

1. \(\vec{r} \cdot \vec{n_1} = x + 3y - z\)
2. \(\vec{r} \cdot \vec{n_2} = 2x - y + z\)

Thus, we have:

\[ x + 3y - z + \lambda(2x - y + z) = 9 + 3\lambda \]

### Step 4: Rearranging the equation.

Rearranging gives:

\[ x + (3 - \lambda)y + (-1 + \lambda)z = 9 + 3\lambda \]

### Step 5: Since the plane passes through the origin, substitute \(x = 0\), \(y = 0\), \(z = 0\).

Substituting these values:

\[ 0 + (3 - \lambda)(0) + (-1 + \lambda)(0) = 9 + 3\lambda \]

This simplifies to:

\[ 0 = 9 + 3\lambda \]

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

From \(0 = 9 + 3\lambda\), we get:

\[ 3\lambda = -9 \implies \lambda = -3 \]

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

Substituting \(\lambda = -3\) back into the equation:

\[ x + (3 - (-3))y + (-1 - 3)z = 9 + 3(-3) \]

This simplifies to:

\[ x + 6y - 4z = 0 \]

### Final Equation of the Plane

Thus, the equation of the plane through the intersection of the given planes and passing through the origin is:

\[ x + 6y - 4z = 0 \]