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

Answer

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

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

### 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:
\[ P = P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (2x + y - z - 3) + \lambda(5x - 3y + 4z - 9) = 0 \]
Expanding this gives:
\[ (2 + 5\lambda)x + (1 - 3\lambda)y + (-1 + 4\lambda)z - (3 + 9\lambda) = 0 \]

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

### Step 4: Find the direction ratios of the plane
The direction ratios of the plane can be obtained from the coefficients of \(x, y, z\) in the plane equation:
- \(a = 2 + 5\lambda\)
- \(b = 1 - 3\lambda\)
- \(c = -1 + 4\lambda\)

### Step 5: Set up the condition for parallelism
For the plane to be parallel to the line, the direction ratios of the line and the normal to the plane must be perpendicular. This means:
\[ (2 + 5\lambda) \cdot 2 + (1 - 3\lambda) \cdot 4 + (-1 + 4\lambda) \cdot 5 = 0 \]

### Step 6: Solve the equation
Expanding the equation:
\[ 4 + 10\lambda + 4 - 12\lambda - 5 + 20\lambda = 0 \]
Combining like terms:
\[ (10\lambda - 12\lambda + 20\lambda) + (4 + 4 - 5) = 0 \]
This simplifies to:
\[ 18\lambda + 3 = 0 \]
Solving for \(\lambda\):
\[ 18\lambda = -3 \quad \Rightarrow \quad \lambda = -\frac{1}{6} \]

### Step 7: Substitute \(\lambda\) back into the plane equation
Substituting \(\lambda = -\frac{1}{6}\) back into the plane equation:
\[ (2 + 5(-\frac{1}{6}))x + (1 - 3(-\frac{1}{6}))y + (-1 + 4(-\frac{1}{6}))z - (3 + 9(-\frac{1}{6})) = 0 \]
Calculating each term:
- For \(x\): \(2 - \frac{5}{6} = \frac{12}{6} - \frac{5}{6} = \frac{7}{6}\)
- For \(y\): \(1 + \frac{1}{2} = \frac{3}{2}\)
- For \(z\): \(-1 - \frac{2}{3} = -\frac{3}{3} - \frac{2}{3} = -\frac{5}{3}\)
- For the constant: \(-3 - \frac{3}{2} = -\frac{6}{2} - \frac{3}{2} = -\frac{9}{2}\)

Thus, the equation becomes:
\[ \frac{7}{6}x + \frac{3}{2}y - \frac{5}{3}z + \frac{9}{2} = 0 \]

### Step 8: Clear the fractions
Multiplying through by 6 to eliminate the fractions:
\[ 7x + 9y - 10z + 27 = 0 \]
Rearranging gives the final equation of the plane:
\[ 7x + 9y - 10z + 27 = 0 \]

### Final Answer
The equation of the required plane is:
\[ 7x + 9y - 10z + 27 = 0 \]