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 and through the origin, we can follow these steps:

### Step 1: Write the equations of the given planes
The equations of the two planes are:
1. Plane P1: \( x + 2y + z - 10 = 0 \)
2. Plane P2: \( 3x + y - z - 5 = 0 \)

### Step 2: Form the equation of the plane passing through the line of intersection
The equation of the plane passing through the line of intersection of the two planes can be expressed in the form:
\[ P1 + \lambda P2 = 0 \]
Substituting the equations of the planes:
\[ (x + 2y + z - 10) + \lambda(3x + y - z - 5) = 0 \]
Expanding this gives:
\[ x + 2y + z - 10 + \lambda(3x + y - z - 5) = 0 \]
This simplifies to:
\[ x + 2y + z - 10 + 3\lambda x + \lambda y - \lambda z - 5\lambda = 0 \]
Combining like terms, we have:
\[ (1 + 3\lambda)x + (2 + \lambda)y + (1 - \lambda)z - (10 + 5\lambda) = 0 \]

### Step 3: Substitute the origin into the equation
Since the plane passes through the origin (0, 0, 0), we substitute \(x = 0\), \(y = 0\), and \(z = 0\) into the equation:
\[ (1 + 3\lambda)(0) + (2 + \lambda)(0) + (1 - \lambda)(0) - (10 + 5\lambda) = 0 \]
This simplifies to:
\[ - (10 + 5\lambda) = 0 \]
From this, we can solve for \(\lambda\):
\[ 10 + 5\lambda = 0 \implies 5\lambda = -10 \implies \lambda = -2 \]

### Step 4: Substitute \(\lambda\) back into the plane equation
Now we substitute \(\lambda = -2\) back into the equation of the plane:
\[ (1 + 3(-2))x + (2 + (-2))y + (1 - (-2))z - (10 + 5(-2)) = 0 \]
This simplifies to:
\[ (1 - 6)x + (2 - 2)y + (1 + 2)z - (10 - 10) = 0 \]
Which further simplifies to:
\[ -5x + 3z = 0 \]
Rearranging gives us:
\[ 5x - 3z = 0 \]

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