Find the equation of the plane passing through the line of intersection of the planes `2x-y=0` and `3z-y=0`, and perpendicular to the plane `4x+5y-3z=9`.

Answer

To find the equation of the plane passing through the line of intersection of the planes \(2x - y = 0\) and \(3z - y = 0\), and perpendicular to the plane \(4x + 5y - 3z = 9\), we can follow these steps:

### Step 1: Identify the equations of the given planes
The equations of the two planes are:
1. Plane 1: \(P_1: 2x - y = 0\)
2. Plane 2: \(P_2: 3z - y = 0\)

### Step 2: Write the equation of the plane through the line of intersection
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:
\[ (2x - y) + \lambda(3z - y) = 0 \]
This simplifies to:
\[ 2x - y + 3\lambda z - \lambda y = 0 \]
Rearranging gives:
\[ 2x + (-(1 + \lambda))y + 3\lambda z = 0 \]

### Step 3: Identify the normal vectors
The normal vector of the plane we just derived is:
\[ \mathbf{n_1} = (2, -(1 + \lambda), 3\lambda) \]
The normal vector of the plane \(4x + 5y - 3z = 9\) is:
\[ \mathbf{n_2} = (4, 5, -3) \]

### Step 4: Set up the perpendicularity condition
For the plane to be perpendicular to \(4x + 5y - 3z = 9\), the dot product of the normal vectors must be zero:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 0 \]
Calculating the dot product:
\[ 2 \cdot 4 + (-(1 + \lambda)) \cdot 5 + 3\lambda \cdot (-3) = 0 \]
This expands to:
\[ 8 - 5(1 + \lambda) - 9\lambda = 0 \]
Simplifying gives:
\[ 8 - 5 - 5\lambda - 9\lambda = 0 \]
\[ 3 - 14\lambda = 0 \]

### Step 5: Solve for \(\lambda\)
Solving for \(\lambda\):
\[ 14\lambda = 3 \implies \lambda = \frac{3}{14} \]

### Step 6: Substitute \(\lambda\) back into the plane equation
Substituting \(\lambda = \frac{3}{14}\) back into the equation of the plane:
\[ 2x + \left(-\left(1 + \frac{3}{14}\right)\right)y + 3\left(\frac{3}{14}\right)z = 0 \]
Calculating the coefficients:
\[ 2x - \left(\frac{14 + 3}{14}\right)y + \frac{9}{14}z = 0 \]
This simplifies to:
\[ 2x - \frac{17}{14}y + \frac{9}{14}z = 0 \]
Multiplying through by 14 to eliminate the fractions:
\[ 28x - 17y + 9z = 0 \]

### Final Equation of the Plane
The equation of the required plane is:
\[ 28x - 17y + 9z = 0 \]

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