Find the equation of the plane through the line of intersection of the planes `x-3y+z+6=0` and `x+2y+3z+5=0`, and passing through the origin.

Answer

To find the equation of the plane that passes through the line of intersection of the planes given by the equations \( P_1: x - 3y + z + 6 = 0 \) and \( P_2: x + 2y + 3z + 5 = 0 \), and also passes through the origin, we can follow these steps:

### Step 1: Write the general equation of the plane
The equation of any plane that passes through the line of intersection of two planes \( P_1 \) and \( P_2 \) can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
where \( \lambda \) is a parameter.

### Step 2: Substitute the equations of the planes
Substituting the equations of \( P_1 \) and \( P_2 \):
\[ (x - 3y + z + 6) + \lambda (x + 2y + 3z + 5) = 0 \]
Expanding this gives:
\[ x - 3y + z + 6 + \lambda x + 2\lambda y + 3\lambda z + 5\lambda = 0 \]

### Step 3: Combine like terms
Combining the coefficients of \( x \), \( y \), \( z \), and the constant term, we get:
\[ (1 + \lambda)x + (-3 + 2\lambda)y + (1 + 3\lambda)z + (6 + 5\lambda) = 0 \]

### Step 4: Use the condition that the plane passes through the origin
Since the plane passes through the origin \((0, 0, 0)\), we substitute \( x = 0 \), \( y = 0 \), and \( z = 0 \) into the equation:
\[ 6 + 5\lambda = 0 \]
Solving for \( \lambda \):
\[ 5\lambda = -6 \implies \lambda = -\frac{6}{5} \]

### Step 5: Substitute \( \lambda \) back into the equation
Now substitute \( \lambda = -\frac{6}{5} \) back into the combined equation:
\[ (1 - \frac{6}{5})x + (-3 + 2(-\frac{6}{5}))y + (1 + 3(-\frac{6}{5}))z = 0 \]
This simplifies to:
\[ -\frac{1}{5}x - \frac{27}{5}y - \frac{13}{5}z = 0 \]

### Step 6: Clear the fractions
To eliminate the fractions, multiply through by \(-5\):
\[ x + 27y + 13z = 0 \]

### Final Answer
The equation of the plane is:
\[ x + 27y + 13z = 0 \]