If `ax+13y+bz+c=0` is a plane through the line intersection of `2x+3y-z+1=0, x+y-2z+3=0` and is perpendicular to the plane `3x-y-2z=4`, then the value of `2a+3b+4c` is equal to

Options

A. `-12`
B. 12
C. 10
D. `-10`

Answer

To solve the problem step by step, we need to find the coefficients \( a \), \( b \), and \( c \) for the plane equation \( ax + 13y + bz + c = 0 \) that passes through the line of intersection of the two given planes and is perpendicular to another given plane.

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

The two planes given are:
1. \( 2x + 3y - z + 1 = 0 \) (let's call this Plane 1)
2. \( x + y - 2z + 3 = 0 \) (let's call this Plane 2)

The equation of a plane through the line of intersection of these two planes can be expressed as:
\[ 2x + 3y - z + 1 + \lambda (x + y - 2z + 3) = 0 \]
where \( \lambda \) is a parameter.

### Step 2: Expand the equation

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

### Step 3: Identify coefficients

From the expanded equation, we can identify the coefficients:
- Coefficient of \( x \): \( a = 2 + \lambda \)
- Coefficient of \( y \): \( b = 3 + \lambda \)
- Coefficient of \( z \): \( c = -1 - 2\lambda \)
- Constant term: \( 1 + 3\lambda \)

### Step 4: Condition for perpendicularity

The plane is also given to be perpendicular to the plane \( 3x - y - 2z = 4 \). The normal vector of this plane is \( (3, -1, -2) \).

For two planes to be perpendicular, the dot product of their normal vectors must be zero. The normal vector of our plane is \( (a, 13, b) \). Therefore, we have:
\[ 3a - 1 \cdot 13 - 2b = 0 \]

### Step 5: Substitute \( a \) and \( b \)

Substituting \( a \) and \( b \) into the equation:
\[ 3(2 + \lambda) - 13 - 2(3 + \lambda) = 0 \]
Expanding this gives:
\[ 6 + 3\lambda - 13 - 6 - 2\lambda = 0 \]
Simplifying this:
\[ \lambda - 13 = 0 \implies \lambda = 7 \]

### Step 6: Substitute \( \lambda \) back to find \( a \), \( b \), and \( c \)

Now substituting \( \lambda = 7 \) back into the equations for \( a \), \( b \), and \( c \):
- \( a = 2 + 7 = 9 \)
- \( b = 3 + 7 = 10 \)
- \( c = -1 - 2(7) = -15 \)

### Step 7: Calculate \( 2a + 3b + 4c \)

Now we need to calculate \( 2a + 3b + 4c \):
\[ 2a + 3b + 4c = 2(9) + 3(10) + 4(-15) \]
Calculating this gives:
\[ = 18 + 30 - 60 = -12 \]

### Final Answer

Thus, the value of \( 2a + 3b + 4c \) is \( -12 \).