The equation of plane passing through the line of intersection of planes `2x-y+z=3,4x-3y-5z+9=0` and parallel to the line `(x+1)/(2)=(y+3)/(4)=(z-3)/(5)` is

Options

A. `11x-3y-2z-54=0`
B. `11x-3y+2z-54=0`
C. `11x+3y-2z-54=0`
D. `11x-3y-2z+54=0`

Answer

To find the equation of the plane passing through the line of intersection of the planes \(2x - y + z = 3\) and \(4x - 3y - 5z + 9 = 0\), and which is parallel to the line given by \(\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z - 3}{5}\), we can follow these steps:

### Step 1: Identify the equations of the planes
The equations of the two planes are:
1. \(P_1: 2x - y + z - 3 = 0\)
2. \(P_2: 4x - 3y - 5z + 9 = 0\)

### Step 2: Form the equation of the plane through the line of intersection
The equation of the plane passing through the line of intersection of the two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (2x - y + z - 3) + \lambda(4x - 3y - 5z + 9) = 0 \]

### Step 3: Expand the equation
Expanding the equation gives:
\[ 2x - y + z - 3 + \lambda(4x - 3y - 5z + 9) = 0 \]
This simplifies to:
\[ (2 + 4\lambda)x + (-1 - 3\lambda)y + (1 - 5\lambda)z + (-3 + 9\lambda) = 0 \]

### Step 4: Identify the direction ratios of the line
The direction ratios of the line given by \(\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z - 3}{5}\) are \(2, 4, 5\).

### Step 5: Set up the condition for parallelism
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 \(z\):
\[ (2 + 4\lambda, -1 - 3\lambda, 1 - 5\lambda) \]
Let \(L_1 = 2 + 4\lambda\), \(M_1 = -1 - 3\lambda\), \(N_1 = 1 - 5\lambda\). The condition for perpendicularity is:
\[ L_1 \cdot 2 + M_1 \cdot 4 + N_1 \cdot 5 = 0 \]

### Step 6: Substitute and simplify
Substituting \(L_1\), \(M_1\), and \(N_1\):
\[ (2 + 4\lambda) \cdot 2 + (-1 - 3\lambda) \cdot 4 + (1 - 5\lambda) \cdot 5 = 0 \]
This expands to:
\[ 4 + 8\lambda - 4 - 12\lambda + 5 - 25\lambda = 0 \]
Combining like terms:
\[ (8\lambda - 12\lambda - 25\lambda) + (4 - 4 + 5) = 0 \]
This simplifies to:
\[ -29\lambda + 5 = 0 \]

### Step 7: Solve for \(\lambda\)
Solving for \(\lambda\):
\[ 29\lambda = 5 \implies \lambda = \frac{5}{29} \]

### Step 8: Substitute \(\lambda\) back into the plane equation
Substituting \(\lambda\) back into the plane equation:
\[ (2 + 4 \cdot \frac{5}{29})x + (-1 - 3 \cdot \frac{5}{29})y + (1 - 5 \cdot \frac{5}{29})z + (-3 + 9 \cdot \frac{5}{29}) = 0 \]
Calculating each term:
- Coefficient of \(x\): \(2 + \frac{20}{29} = \frac{58 + 20}{29} = \frac{78}{29}\)
- Coefficient of \(y\): \(-1 - \frac{15}{29} = -\frac{29 + 15}{29} = -\frac{44}{29}\)
- Coefficient of \(z\): \(1 - \frac{25}{29} = \frac{29 - 25}{29} = \frac{4}{29}\)
- Constant term: \(-3 + \frac{45}{29} = -\frac{87 + 45}{29} = -\frac{42}{29}\)

### Step 9: Final equation of the plane
Putting it all together, we have:
\[ \frac{78}{29}x - \frac{44}{29}y + \frac{4}{29}z - \frac{42}{29} = 0 \]
Multiplying through by \(29\) to eliminate the fraction:
\[ 78x - 44y + 4z - 42 = 0 \]
This can be simplified to:
\[ 39x - 22y + 2z = 21 \]

### Final Answer
The equation of the plane is:
\[ 39x - 22y + 2z = 21 \]