Find the equation of the plane passing through the intersection of the planes `2x+2y-3z-7=0 and 2x+5y+3z-9=0` such that the intercepts made by the resulting plane one the x-axis and the z-axis are equal.

Answer

To find the equation of the plane passing through the intersection of the planes \(2x + 2y - 3z - 7 = 0\) and \(2x + 5y + 3z - 9 = 0\) such that the intercepts made by the resulting plane on the x-axis and the z-axis are equal, we can follow these steps:

### Step 1: Write the equations of the given planes
The equations of the two planes are:
1. \(P_1: 2x + 2y - 3z - 7 = 0\)
2. \(P_2: 2x + 5y + 3z - 9 = 0\)

### Step 2: Form the equation of the resulting plane
The equation of any plane passing through the intersection of these two planes can be expressed as:
\[ P = P_1 + \lambda P_2 \]
Substituting \(P_1\) and \(P_2\):
\[ P = (2x + 2y - 3z - 7) + \lambda(2x + 5y + 3z - 9) \]
This simplifies to:
\[ P = (2 + 2\lambda)x + (2 + 5\lambda)y + (-3 + 3\lambda)z - (7 + 9\lambda) = 0 \]

### Step 3: Identify the coefficients
From the equation, we can identify the coefficients:
- Coefficient of \(x\): \(A = 2 + 2\lambda\)
- Coefficient of \(y\): \(B = 2 + 5\lambda\)
- Coefficient of \(z\): \(C = -3 + 3\lambda\)
- Constant term: \(D = -(7 + 9\lambda)\)

### Step 4: Find the intercepts
The intercepts on the axes can be found using the formula:
- Intercept on the x-axis: \(x = -\frac{D}{A} = \frac{7 + 9\lambda}{2 + 2\lambda}\)
- Intercept on the z-axis: \(z = -\frac{D}{C} = \frac{7 + 9\lambda}{-3 + 3\lambda}\)

### Step 5: Set the intercepts equal
According to the problem, the intercepts on the x-axis and z-axis are equal:
\[ \frac{7 + 9\lambda}{2 + 2\lambda} = \frac{7 + 9\lambda}{-3 + 3\lambda} \]
Cross-multiplying gives:
\[ (7 + 9\lambda)(-3 + 3\lambda) = (7 + 9\lambda)(2 + 2\lambda) \]

### Step 6: Solve for \(\lambda\)
Assuming \(7 + 9\lambda \neq 0\), we can divide both sides by \(7 + 9\lambda\):
\[ -3 + 3\lambda = 2 + 2\lambda \]
Rearranging gives:
\[ 3\lambda - 2\lambda = 2 + 3 \]
\[ \lambda = 5 \]

### Step 7: Substitute \(\lambda\) back into the plane equation
Substituting \(\lambda = 5\) back into the equation of the plane:
\[ P = (2 + 2(5))x + (2 + 5(5))y + (-3 + 3(5))z - (7 + 9(5)) = 0 \]
This simplifies to:
\[ 12x + 27y + 12z - 50 = 0 \]

### Final Equation
Rearranging gives the final equation of the plane:
\[ 12x + 27y + 12z = 50 \]