Let two planes `p _(1): 2x -y + z =2 and p _(2) : x + 2y - z=3` are given :
equation of the plane through the intersection of `p _(1) and p_(2)` and the point `(3,2,1)` is :

Options

A. `3x -y + 2z -9=0`
B. `x-3y+ 2z +1 =0`
C. `2x - 3y + z-1 =0`
D. `4x - 3y + 2z -8=0`

Answer

To find the equation of the plane that passes through the intersection of the two given planes \( P_1: 2x - y + z = 2 \) and \( P_2: x + 2y - z = 3 \) and also passes through the point \( (3, 2, 1) \), we can follow these steps:

### Step 1: Write the equation of the plane through the intersection of the two planes.

The equation of a plane passing through the intersection of the two planes can be expressed as:
\[ P = P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes \( P_1 \) and \( P_2 \):
\[ (2x - y + z - 2) + \lambda (x + 2y - z - 3) = 0 \]

### Step 2: Expand the equation.

Expanding the equation gives:
\[ 2x - y + z - 2 + \lambda x + 2\lambda y - \lambda z - 3\lambda = 0 \]
Combining like terms results in:
\[ (2 + \lambda)x + (-1 + 2\lambda)y + (1 - \lambda)z - (2 + 3\lambda) = 0 \]

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

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

### Step 4: Simplify the equation.

Expanding this gives:
\[ 3(2 + \lambda) + 2(-1 + 2\lambda) + (1 - \lambda) - (2 + 3\lambda) = 0 \]
This simplifies to:
\[ 6 + 3\lambda - 2 + 4\lambda + 1 - \lambda - 2 - 3\lambda = 0 \]
Combining like terms results in:
\[ (6 - 2 + 1 - 2) + (3\lambda + 4\lambda - \lambda - 3\lambda) = 0 \]
This simplifies to:
\[ 3 + 3\lambda = 0 \]

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

Setting the equation to zero:
\[ 3 + 3\lambda = 0 \implies 3\lambda = -3 \implies \lambda = -1 \]

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

Now substitute \( \lambda = -1 \) back into the equation of the plane:
\[ (2 - 1)x + (-1 - 2)y + (1 + 1)z - (2 - 3) = 0 \]
This simplifies to:
\[ 1x - 3y + 2z + 1 = 0 \]
Thus, the equation of the required plane is:
\[ x - 3y + 2z + 1 = 0 \]

### Final Answer:
The equation of the plane through the intersection of the given planes and the point \( (3, 2, 1) \) is:
\[ x - 3y + 2z + 1 = 0 \]