Find the equation of plane through the line of intersection of the planes 3x – 4y + 5z = 10, 2x +2y-3z=4 and parallel to the line x=2y=3z

Answer

To find the equation of the plane through the line of intersection of the planes \(3x - 4y + 5z = 10\) and \(2x + 2y - 3z = 4\), and parallel to the line given by \(x = 2y = 3z\), we can follow these steps:

### Step 1: Write the equation of the plane through the line of intersection
The equation of a plane that passes through the line of intersection of two planes can be expressed as:
\[ 3x - 4y + 5z - 10 + \lambda(2x + 2y - 3z - 4) = 0 \]
where \(\lambda\) is a parameter.

### Step 2: Simplify the equation
Expanding the equation gives:
\[ (3 + 2\lambda)x + (-4 + 2\lambda)y + (5 - 3\lambda)z - (10 + 4\lambda) = 0 \]

### Step 3: Identify the direction ratios of the given line
The line \(x = 2y = 3z\) can be expressed in parametric form as:
\[ x = 6t, \quad y = 3t, \quad z = 2t \]
From this, we can see that the direction ratios of the line are \(6, 3, 2\).

### Step 4: Set up the normal condition
For the plane to be parallel to the line, the normal vector of the plane must be perpendicular to the direction ratios of the line. The normal vector of the plane is given by the coefficients of \(x\), \(y\), and \(z\):
\[ (3 + 2\lambda, -4 + 2\lambda, 5 - 3\lambda) \]
We need to ensure that this normal vector is perpendicular to the direction ratios \(6, 3, 2\). Therefore, we set up the dot product:
\[ (3 + 2\lambda) \cdot 6 + (-4 + 2\lambda) \cdot 3 + (5 - 3\lambda) \cdot 2 = 0 \]

### Step 5: Solve the equation
Expanding the dot product gives:
\[ 18 + 12\lambda - 12 + 6\lambda + 10 - 6\lambda = 0 \]
Combining like terms results in:
\[ 18 + 12\lambda - 12 + 10 - 6\lambda = 0 \]
This simplifies to:
\[ 12 + 6\lambda = 0 \]
Thus,
\[ 6\lambda = -12 \implies \lambda = -2 \]

### Step 6: Substitute \(\lambda\) back into the plane equation
Now substituting \(\lambda = -2\) back into the plane equation:
\[ (3 + 2(-2))x + (-4 + 2(-2))y + (5 - 3(-2))z - (10 + 4(-2)) = 0 \]
This simplifies to:
\[ (-1)x + (-8)y + (11)z - 2 = 0 \]
or rearranging gives:
\[ x + 8y - 11z = 2 \]

### Final Equation of the Plane
Thus, the equation of the required plane is:
\[ x + 8y - 11z = 2 \]