Find the equation of the plane passing through the intersection of the planes: `x+y+z+1=0 and 2x-3y +5z-2=0` and the point `(-1, 2,1)`

Answer

To find the equation of the plane passing through the intersection of the planes \(x + y + z + 1 = 0\) and \(2x - 3y + 5z - 2 = 0\), and also passing through the point \((-1, 2, 1)\), we can follow these steps:

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

For our planes:
- \(P_1: x + y + z + 1 = 0\)
- \(P_2: 2x - 3y + 5z - 2 = 0\)

Thus, the equation of the required plane becomes:
\[ (x + y + z + 1) + \lambda(2x - 3y + 5z - 2) = 0 \]

### Step 2: Substitute the point \((-1, 2, 1)\) into the equation
We need to find the value of \(\lambda\) such that the plane passes through the point \((-1, 2, 1)\). Substitute \(x = -1\), \(y = 2\), and \(z = 1\) into the equation:
\[ (-1 + 2 + 1 + 1) + \lambda(2(-1) - 3(2) + 5(1) - 2) = 0 \]

### Step 3: Simplify the equation
First, simplify the left side:
\[ (-1 + 2 + 1 + 1) = 3 \]
Now simplify the term involving \(\lambda\):
\[ 2(-1) - 3(2) + 5(1) - 2 = -2 - 6 + 5 - 2 = -5 \]
So we have:
\[ 3 + \lambda(-5) = 0 \]

### Step 4: Solve for \(\lambda\)
Rearranging gives:
\[ 3 - 5\lambda = 0 \implies 5\lambda = 3 \implies \lambda = \frac{3}{5} \]

### Step 5: Substitute \(\lambda\) back into the plane equation
Now substitute \(\lambda = \frac{3}{5}\) back into the equation of the plane:
\[ x + y + z + 1 + \frac{3}{5}(2x - 3y + 5z - 2) = 0 \]
Distributing \(\frac{3}{5}\):
\[ x + y + z + 1 + \left(\frac{6}{5}x - \frac{9}{5}y + 3z - \frac{6}{5}\right) = 0 \]

### Step 6: Combine like terms
Combine the terms:
\[ \left(1 + \frac{6}{5}\right)x + \left(1 - \frac{9}{5}\right)y + \left(1 + 3\right)z + \left(1 - \frac{6}{5}\right) = 0 \]
This simplifies to:
\[ \frac{11}{5}x - \frac{4}{5}y + 4z - \frac{1}{5} = 0 \]
Multiplying through by 5 to eliminate the fractions gives:
\[ 11x - 4y + 20z - 1 = 0 \]

### Final Equation
Thus, the equation of the required plane is:
\[ 11x - 4y + 20z - 1 = 0 \]