Let the plane passing through the point `(-1,0,-2)` and perpendicular to each of the planes `2x+y-z=2 and x-y-z=3` be `ax+by+cz+8=0`. Then the value of `a+b+c` is equal to

Options

A. 3
B. 8
C. 5
D. 4

Answer

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) for the plane equation \( ax + by + cz + 8 = 0 \) that passes through the point \((-1, 0, -2)\) and is perpendicular to the given planes \( 2x + y - z = 2 \) and \( x - y - z = 3 \).

### Step 1: Substitute the point into the plane equation

The plane passes through the point \((-1, 0, -2)\). We substitute these coordinates into the plane equation:

\[ a(-1) + b(0) + c(-2) + 8 = 0 \]

This simplifies to:

\[ -a - 2c + 8 = 0 \]

Rearranging gives us our first equation:

\[ a + 2c = 8 \quad \text{(Equation 1)} \]

### Step 2: Find the normal vectors of the given planes

The normal vector of the first plane \( 2x + y - z = 2 \) is \( \vec{n_1} = (2, 1, -1) \).

The normal vector of the second plane \( x - y - z = 3 \) is \( \vec{n_2} = (1, -1, -1) \).

### Step 3: Find the normal vector of the required plane

Since the required plane is perpendicular to both given planes, its normal vector \( \vec{n} = (a, b, c) \) must be orthogonal to both \( \vec{n_1} \) and \( \vec{n_2} \). This gives us two equations:

1. The dot product with \( \vec{n_1} \):
\[ 2a + b - c = 0 \quad \text{(Equation 2)} \]

2. The dot product with \( \vec{n_2} \):
\[ a - b - c = 0 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations

Now we have three equations to solve:

1. \( a + 2c = 8 \) (Equation 1)
2. \( 2a + b - c = 0 \) (Equation 2)
3. \( a - b - c = 0 \) (Equation 3)

We can solve these equations step by step.

From Equation 3, we can express \( b \) in terms of \( a \) and \( c \):

\[ b = a - c \quad \text{(Substituting into Equation 2)} \]

Substituting \( b \) into Equation 2:

\[ 2a + (a - c) - c = 0 \]
\[ 3a - 2c = 0 \quad \Rightarrow \quad 3a = 2c \quad \Rightarrow \quad c = \frac{3}{2}a \]

### Step 5: Substitute \( c \) back into Equation 1

Now substitute \( c = \frac{3}{2}a \) into Equation 1:

\[ a + 2\left(\frac{3}{2}a\right) = 8 \]
\[ a + 3a = 8 \]
\[ 4a = 8 \quad \Rightarrow \quad a = 2 \]

### Step 6: Find \( c \) and \( b \)

Now, substituting \( a = 2 \) back into \( c = \frac{3}{2}a \):

\[ c = \frac{3}{2}(2) = 3 \]

Now substitute \( a \) into \( b = a - c \):

\[ b = 2 - 3 = -1 \]

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

Now we have:

- \( a = 2 \)
- \( b = -1 \)
- \( c = 3 \)

Thus,

\[ a + b + c = 2 - 1 + 3 = 4 \]

### Final Answer

The value of \( a + b + c \) is \( \boxed{4} \).