A plane P = 0 passes through the line of intersection of the planes `x+y+z+3=0 and x-y+z-2=0.` If the plane P divides the ratio `2:1` internally and the equation of the plane is `ax-2y+bz=c` where `a, b, c in N`, then the value of `3a+4b-5c` is equal to

Options

A. 22
B. 32
C. 42
D. 10

Answer

To solve the problem step by step, we need to find the equation of the plane \( P = 0 \) that passes through the line of intersection of the given planes and divides the line in the ratio \( 2:1 \).

### Step 1: Find the line of intersection of the given planes

The equations of the planes are:
1. \( x + y + z + 3 = 0 \) (Plane 1)
2. \( x - y + z - 2 = 0 \) (Plane 2)

To find the line of intersection, we can express one variable in terms of the others. Let's eliminate \( z \) from both equations.

From Plane 1:
\[ z = -x - y - 3 \]

Substituting \( z \) in Plane 2:
\[ x - y + (-x - y - 3) - 2 = 0 \]
This simplifies to:
\[ -x - 2y - 5 = 0 \implies x + 2y = -5 \implies x = -5 - 2y \]

Now substituting \( x \) back into the expression for \( z \):
\[ z = -(-5 - 2y) - y - 3 = 5 + 2y - y - 3 = 2 + y \]

Thus, we can express the line of intersection in parametric form:
Let \( y = t \):
\[ x = -5 - 2t, \quad y = t, \quad z = 2 + t \]

### Step 2: Find the equation of the plane \( P \)

The plane \( P \) can be expressed as:
\[ P: x + y + z + 3 + \lambda (x - y + z - 2) = 0 \]
Expanding this gives:
\[ (1 + \lambda)x + (1 - \lambda)y + (1 + \lambda)z + (3 - 2\lambda) = 0 \]

### Step 3: Determine the value of \( \lambda \)

The plane \( P \) divides the line of intersection in the ratio \( 2:1 \). To find the specific point on the line, we can use the section formula. The coordinates of the point dividing the segment in the ratio \( 2:1 \) are given by:
\[ \left( \frac{2(-5) + 1(-5)}{2 + 1}, \frac{2(0) + 1(1)}{2 + 1}, \frac{2(2) + 1(2)}{2 + 1} \right) \]
Calculating this gives:
\[ \left( \frac{-10 - 5}{3}, \frac{0 + 1}{3}, \frac{4 + 2}{3} \right) = \left( -\frac{15}{3}, \frac{1}{3}, \frac{6}{3} \right) = (-5, \frac{1}{3}, 2) \]

### Step 4: Substitute the point into the plane equation

Substituting \( (-5, \frac{1}{3}, 2) \) into the plane equation:
\[ (1 + \lambda)(-5) + (1 - \lambda)\left(\frac{1}{3}\right) + (1 + \lambda)(2) + (3 - 2\lambda) = 0 \]
Simplifying this:
\[ -5 - 5\lambda + \frac{1}{3} - \frac{1}{3}\lambda + 2 + 2\lambda + 3 - 2\lambda = 0 \]
Combining like terms:
\[ 0 + (-5\lambda - \frac{1}{3}\lambda + 2\lambda - 2\lambda) + \left(-5 + \frac{1}{3} + 2 + 3\right) = 0 \]
This leads to:
\[ \frac{-15\lambda - \lambda + 6}{3} = 0 \]
Solving for \( \lambda \):
\[ -16\lambda + 6 = 0 \implies \lambda = \frac{6}{16} = \frac{3}{8} \]

### Step 5: Substitute \( \lambda \) back into the plane equation

Substituting \( \lambda \) back into the plane equation gives:
\[ (1 + \frac{3}{8})x + (1 - \frac{3}{8})y + (1 + \frac{3}{8})z + (3 - 2 \cdot \frac{3}{8}) = 0 \]
This simplifies to:
\[ \frac{11}{8}x + \frac{5}{8}y + \frac{11}{8}z + \frac{21}{8} = 0 \]
Multiplying through by \( 8 \) gives:
\[ 11x + 5y + 11z + 21 = 0 \]

### Step 6: Compare with the given form

The equation of the plane is given as \( ax - 2y + bz = c \). Comparing:
\[ 11x - 2y + 11z + 21 = 0 \]
We identify:
- \( a = 11 \)
- \( b = 11 \)
- \( c = -21 \)

### Step 7: Calculate \( 3a + 4b - 5c \)

Now we calculate:
\[ 3a + 4b - 5c = 3(11) + 4(11) - 5(-21) \]
Calculating each term:
\[ = 33 + 44 + 105 = 182 \]

### Final Answer

Thus, the value of \( 3a + 4b - 5c \) is \( 182 \).