The equation of a plane passing through the line of intersection of the planes :
`x+2y+z-10=0 and 3x+y-z=5` and passing through the origin is :

Options

A. `5x+3z=0`
B. `5x-3z=0`
C. `5x+4y+3z=0`
D. `5x-4y+3z=0`

Answer

To find the equation of the plane passing through the line of intersection of the given planes \(x + 2y + z - 10 = 0\) and \(3x + y - z - 5 = 0\), and also passing through the origin, we can follow these steps:

### Step 1: Identify the equations of the planes
We have two planes given by the equations:
1. \(P_1: x + 2y + z - 10 = 0\)
2. \(P_2: 3x + y - z - 5 = 0\)

### Step 2: Write the general equation of the plane
The equation of a plane passing through the line of intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes, we get:
\[ (x + 2y + z - 10) + \lambda(3x + y - z - 5) = 0 \]

### Step 3: Expand the equation
Expanding the equation gives:
\[ x + 2y + z - 10 + \lambda(3x + y - z - 5) = 0 \]
This can be rewritten as:
\[ (1 + 3\lambda)x + (2 + \lambda)y + (1 - \lambda)z - (10 + 5\lambda) = 0 \]

### Step 4: Substitute the point (0, 0, 0)
Since the plane passes through the origin \((0, 0, 0)\), we substitute \(x = 0\), \(y = 0\), and \(z = 0\):
\[ 0 + 0 + 0 - (10 + 5\lambda) = 0 \]
This simplifies to:
\[ 10 + 5\lambda = 0 \]
Solving for \(\lambda\):
\[ 5\lambda = -10 \implies \lambda = -2 \]

### Step 5: Substitute \(\lambda\) back into the equation
Now, substitute \(\lambda = -2\) back into the expanded equation:
\[ (1 + 3(-2))x + (2 + (-2))y + (1 - (-2))z - (10 + 5(-2)) = 0 \]
This simplifies to:
\[ (1 - 6)x + 0y + (1 + 2)z - (10 - 10) = 0 \]
Which further simplifies to:
\[ -5x + 3z = 0 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 5x - 3z = 0 \]

### Final Answer
Thus, the equation of the plane is:
\[ 5x - 3z = 0 \]

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