Find the vector equation of a plane containing line of intersection of plane `vecr cdot(2hati -hatj-hatk)=6`
and plane `vecr cdot (3hati +hatj +2hatk)=0` and passing through (-1, 1, 2).

Answer

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

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

The first plane is given by:
\[ \vec{r} \cdot (2\hat{i} - \hat{j} - \hat{k}) = 6 \]
This can be rewritten in Cartesian form as:
\[ 2x - y - z - 6 = 0 \quad \text{(Equation 1)} \]

The second plane is given by:
\[ \vec{r} \cdot (3\hat{i} + \hat{j} + 2\hat{k}) = 0 \]
This can be rewritten in Cartesian form as:
\[ 3x + y + 2z = 0 \quad \text{(Equation 2)} \]

### Step 2: Form the general equation of the plane containing the line of intersection.

The general equation of a plane containing the line of intersection of the two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes into this form gives:
\[ (2x - y - z - 6) + \lambda(3x + y + 2z) = 0 \]

### Step 3: Simplify the equation.

Expanding this, we have:
\[ (2 + 3\lambda)x + (-1 + \lambda)y + (-1 + 2\lambda)z - 6 = 0 \]

### Step 4: Substitute the point (-1, 1, 2) into the equation.

Since the plane passes through the point (-1, 1, 2), we substitute \(x = -1\), \(y = 1\), and \(z = 2\) into the equation:
\[ (2 + 3\lambda)(-1) + (-1 + \lambda)(1) + (-1 + 2\lambda)(2) - 6 = 0 \]

### Step 5: Solve for \(\lambda\).

This simplifies to:
\[ -(2 + 3\lambda) - 1 + \lambda - 2 + 4\lambda - 6 = 0 \]
Combining like terms:
\[ -2 - 3\lambda - 1 + \lambda - 2 + 4\lambda - 6 = 0 \]
\[ -11 + 2\lambda = 0 \]
Thus,
\[ 2\lambda = 11 \implies \lambda = \frac{11}{2} \]

### Step 6: Substitute \(\lambda\) back into the plane equation.

Now substitute \(\lambda = \frac{11}{2}\) back into the plane equation:
\[ (2 + 3 \cdot \frac{11}{2})x + (-1 + \frac{11}{2})y + (-1 + 2 \cdot \frac{11}{2})z - 6 = 0 \]
Calculating the coefficients:
\[ (2 + \frac{33}{2})x + (\frac{9}{2})y + (10)z - 6 = 0 \]
This simplifies to:
\[ \frac{37}{2}x + \frac{9}{2}y + 10z - 6 = 0 \]

### Step 7: Multiply through by 2 to eliminate the fractions.

Multiplying through by 2 gives:
\[ 37x + 9y + 20z - 12 = 0 \]

### Final Answer:

The vector equation of the required plane is:
\[ 37x + 9y + 20z = 12 \]