Let the plane P passes througn the point (1,2,3) and it contains the line of intersection of `vecr*(hati+hatj+4hatk)=16 "and" vecr*(-hati+hatj+hatk)=6` . Then which of the following point does not lie on P?

Options

A. (-8,8,6)
B. (-4,3,5)
C. (8,-5,1)
D. (-8,8,5)

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the equations of the given planes
The equations of the two planes provided are:
1. \( \vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 16 \)
2. \( \vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 6 \)

We can rewrite these equations in Cartesian form:
1. \( x + y + 4z = 16 \) (Equation of Plane 1)
2. \( -x + y + z = 6 \) (Equation of Plane 2)

### Step 2: Formulate the equation of the new plane
The new plane \( P \) passes through the point \( (1, 2, 3) \) and contains the line of intersection of the two planes. The equation of the new plane can be expressed as:
\[ P: P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes, we have:
\[ (x + y + 4z - 16) + \lambda (-x + y + z - 6) = 0 \]

### Step 3: Substitute the point into the plane equation
We substitute the point \( (1, 2, 3) \) into the equation to find the value of \( \lambda \):
\[ (1 + 2 + 4 \cdot 3 - 16) + \lambda (-1 + 2 + 3 - 6) = 0 \]
Calculating the left-hand side:
\[ (1 + 2 + 12 - 16) + \lambda (-2) = 0 \]
This simplifies to:
\[ -1 - 2\lambda = 0 \]
From this, we find:
\[ \lambda = -\frac{1}{2} \]

### Step 4: Substitute \( \lambda \) back into the plane equation
Now we substitute \( \lambda = -\frac{1}{2} \) back into the equation of the plane \( P \):
\[ P: (x + y + 4z - 16) - \frac{1}{2}(-x + y + z - 6) = 0 \]
This simplifies to:
\[ x + y + 4z - 16 + \frac{1}{2}x - \frac{1}{2}y - \frac{1}{2}z + 3 = 0 \]
Combining like terms:
\[ \frac{3}{2}x + \frac{1}{2}y + \frac{7}{2}z - 13 = 0 \]
Multiplying through by 2 to eliminate the fractions:
\[ 3x + y + 7z - 26 = 0 \]
Thus, the equation of the plane \( P \) is:
\[ 3x + y + 7z = 26 \]

### Step 5: Check which points lie on the plane
We will check each of the given points against the plane equation \( 3x + y + 7z = 26 \):

1. **Point A: \( (-8, 8, 6) \)**
\[ 3(-8) + 8 + 7(6) = -24 + 8 + 42 = 26 \quad \text{(lies on P)} \]

2. **Point B: \( (-4, 3, 5) \)**
\[ 3(-4) + 3 + 7(5) = -12 + 3 + 35 = 26 \quad \text{(lies on P)} \]

3. **Point C: \( (8, -5, 1) \)**
\[ 3(8) - 5 + 7(1) = 24 - 5 + 7 = 26 \quad \text{(lies on P)} \]

4. **Point D: \( (-8, 8, 5) \)**
\[ 3(-8) + 8 + 7(5) = -24 + 8 + 35 = 19 \quad \text{(does not lie on P)} \]

### Conclusion
The point that does not lie on the plane \( P \) is \( (-8, 8, 5) \).

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