An equation of the plane passing through the origin and containing the lines whose direction cosinesare proportional to `1, -2,2` and `2, 3, -1` is

Options

A. A) x - 2y + 2z = 0
B. b) 2x + 3y - z = 0
C. C ) x + 5y - 3z = 0
D. D) 4x - 5y - 7z = 0

Answer

To find the equation of the plane passing through the origin and containing the lines whose direction cosines are proportional to \(1, -2, 2\) and \(2, 3, -1\), we can follow these steps:

### Step 1: Understand the Direction Cosines
The direction cosines of the lines are given as:
- For the first line: \(l_1 = 1\), \(m_1 = -2\), \(n_1 = 2\)
- For the second line: \(l_2 = 2\), \(m_2 = 3\), \(n_2 = -1\)

### Step 2: Set Up the Equation of the Plane
The general equation of a plane passing through the origin can be expressed as:
\[ Lx + My + Nz = 0 \]
where \(L\), \(M\), and \(N\) are the direction cosines of the normal to the plane.

### Step 3: Use the Perpendicularity Condition
Since the plane contains the two lines, the direction ratios of the lines must be perpendicular to the normal of the plane. This gives us two equations based on the dot product being zero.

1. For the first line:
\[ L \cdot 1 + M \cdot (-2) + N \cdot 2 = 0 \quad \text{(Equation 1)} \]
This simplifies to:
\[ L - 2M + 2N = 0 \]

2. For the second line:
\[ L \cdot 2 + M \cdot 3 + N \cdot (-1) = 0 \quad \text{(Equation 2)} \]
This simplifies to:
\[ 2L + 3M - N = 0 \]

### Step 4: Solve the System of Equations
Now we have a system of two equations:
1. \(L - 2M + 2N = 0\)
2. \(2L + 3M - N = 0\)

We can solve these equations simultaneously.

From Equation 1, we can express \(L\) in terms of \(M\) and \(N\):
\[ L = 2M - 2N \quad \text{(Equation 3)} \]

Substituting Equation 3 into Equation 2:
\[ 2(2M - 2N) + 3M - N = 0 \]
This simplifies to:
\[ 4M - 4N + 3M - N = 0 \]
Combining like terms:
\[ 7M - 5N = 0 \]
Thus, we can express \(M\) in terms of \(N\):
\[ M = \frac{5}{7}N \quad \text{(Equation 4)} \]

### Step 5: Substitute Back to Find \(L\), \(M\), and \(N\)
Now substituting Equation 4 into Equation 3:
\[ L = 2\left(\frac{5}{7}N\right) - 2N = \frac{10}{7}N - 2N = \frac{10}{7}N - \frac{14}{7}N = -\frac{4}{7}N \]

### Step 6: Choose a Value for \(N\)
Let’s choose \(N = 7\) (to eliminate the fraction):
- Then \(M = \frac{5}{7} \cdot 7 = 5\)
- And \(L = -\frac{4}{7} \cdot 7 = -4\)

### Step 7: Write the Equation of the Plane
Substituting \(L\), \(M\), and \(N\) back into the plane equation:
\[ -4x + 5y - 7z = 0 \]
or rearranging gives:
\[ 4x - 5y + 7z = 0 \]

### Final Answer
The equation of the plane is:
\[ 4x - 5y + 7z = 0 \]