Given the equation of the line `3x-y+z+1=0 and 5x-y+3z=0`. Then,which of the following is correct?

Options

A. Symmetrical form of the equation of line is `(x)/(2)=(y-(1)/(8))/(-1)=(z+(5)/(8))/(1)`.
B. Symmetrical form of the equation of line is `(x+(1)/(8))/(1)=(y-(5)/(8))/(-1)=(z)/(-2)`
C. Equation of the through `(2, 1, 4)` and perpencular to the given lines is `2x-y+z-7=0`.
D. Equation of the plane through `(2, 1, 4)` and perpendicular to the given lines is `x+y-2z+5=0`.

Answer

To solve the problem step by step, we will analyze the given equations of the lines and derive the necessary conclusions.

### Step 1: Write down the equations of the lines
The equations given are:
1. \( 3x - y + z + 1 = 0 \) (Equation 1)
2. \( 5x - y + 3z = 0 \) (Equation 2)

### Step 2: Rearranging the equations
We can rearrange both equations to express \( z \) in terms of \( x \) and \( y \).

From Equation 1:
\[ z = -3x + y - 1 \]

From Equation 2:
\[ 3z = -5x + y \]
\[ z = \frac{-5x + y}{3} \]

### Step 3: Set the equations for \( z \) equal to each other
Now, we can set the two expressions for \( z \) equal to find a relationship between \( x \) and \( y \):
\[ -3x + y - 1 = \frac{-5x + y}{3} \]

### Step 4: Clear the fraction
Multiply through by 3 to eliminate the fraction:
\[ 3(-3x + y - 1) = -5x + y \]
\[ -9x + 3y - 3 = -5x + y \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ -9x + 5x + 3y - y - 3 = 0 \]
\[ -4x + 2y - 3 = 0 \]
\[ 2y = 4x + 3 \]
\[ y = 2x + \frac{3}{2} \]

### Step 6: Substitute back to find \( z \)
Now substitute \( y \) back into one of the equations for \( z \):
Using \( z = -3x + y - 1 \):
\[ z = -3x + (2x + \frac{3}{2}) - 1 \]
\[ z = -3x + 2x + \frac{3}{2} - 1 \]
\[ z = -x + \frac{1}{2} \]

### Step 7: Write the parametric equations
Now we can express the line in parametric form:
Let \( x = t \):
\[ y = 2t + \frac{3}{2} \]
\[ z = -t + \frac{1}{2} \]

### Step 8: Symmetrical form of the line
From the parametric equations, we can write the symmetrical form:
\[ \frac{x - 0}{1} = \frac{y - \frac{3}{2}}{2} = \frac{z - \frac{1}{2}}{-1} \]

### Step 9: Finding the equation of the plane
To find the equation of the plane that passes through the point \( (2, 1, 4) \) and is perpendicular to the line, we need the direction ratios of the line, which are \( (1, 2, -1) \).

Let the equation of the plane be:
\[ a(x - 2) + b(y - 1) + c(z - 4) = 0 \]
Where \( a, b, c \) are the direction ratios of the line.

### Step 10: Set up the equations for perpendicularity
The plane is perpendicular to the line, so:
1. \( 3a - b + c = 0 \)
2. \( 5a + b + 3c = 0 \)

### Step 11: Solve for \( a, b, c \)
From the equations, we can express \( a, b, c \) in terms of a parameter \( k \):
Let \( a = -4k, b = -4k, c = 8k \).

### Step 12: Substitute back into the plane equation
Substituting these values back into the plane equation gives:
\[ -4k(x - 2) - 4k(y - 1) + 8k(z - 4) = 0 \]
Factoring out \( -4k \):
\[ x - 2 + y - 1 - 2z + 8 = 0 \]
This simplifies to:
\[ x + y - 2z + 5 = 0 \]

### Conclusion
Thus, the correct answer is option D: \( x + y - 2z + 5 = 0 \).