Find the equation of the line passing through the point (1, 2, 3) and perpendicular to the lines
`(x-1)/(3)=(y-3)/(1)=(z-4)/(-5)and(x-1)/(1)=(y-2)/(2)=(z-3)/(3)`

Options

A. `(x-1)/(13)=(y-2)/(-14)=(z-3)/(5)`
B. `(x-1)/(3)=(y+5)/(-8)=(z-3)/(5)and(x-1)/(1)=(y-2)/(2)=(z-3)/(3)`
C. `(x-1)/(3)=(y+5)/(-8)=(z+2)/(5)`
D. `(x-1)/(3)=(y-2)/(-8)=(z-3)/(5)`

Answer

To find the equation of the line passing through the point (1, 2, 3) and perpendicular to the given lines, we can follow these steps:

### Step 1: Identify the direction ratios of the given lines
The first line is given by the equation:
\[ \frac{x-1}{3} = \frac{y-3}{1} = \frac{z-4}{-5} \]
From this, we can extract the direction ratios \( \mathbf{a} = (3, 1, -5) \).

The second line is given by the equation:
\[ \frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3} \]
From this, we can extract the direction ratios \( \mathbf{b} = (1, 2, 3) \).

### Step 2: Find the cross product of the direction ratios
To find a vector that is perpendicular to both lines, we calculate the cross product \( \mathbf{a} \times \mathbf{b} \).

\[ \mathbf{a} = (3, 1, -5), \quad \mathbf{b} = (1, 2, 3) \]

The cross product is calculated using the determinant of a matrix:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -5 \\ 1 & 2 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 1 & -5 \\ 2 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & -5 \\ 1 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 1 & -5 \\ 2 & 3 \end{vmatrix} = (1)(3) - (-5)(2) = 3 + 10 = 13 \)

2. \( \begin{vmatrix} 3 & -5 \\ 1 & 3 \end{vmatrix} = (3)(3) - (-5)(1) = 9 + 5 = 14 \)

3. \( \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = (3)(2) - (1)(1) = 6 - 1 = 5 \)

Putting it all together:
\[ \mathbf{a} \times \mathbf{b} = (13) \mathbf{i} - (14) \mathbf{j} + (5) \mathbf{k} = (13, -14, 5) \]

### Step 3: Write the equation of the line
The direction ratios of the line we are looking for are \( (13, -14, 5) \) and it passes through the point \( (1, 2, 3) \).

The equation of the line in symmetric form is given by:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \]
Substituting \( (x_1, y_1, z_1) = (1, 2, 3) \) and \( (a, b, c) = (13, -14, 5) \):
\[ \frac{x - 1}{13} = \frac{y - 2}{-14} = \frac{z - 3}{5} \]

### Final Answer
The equation of the line is:
\[ \frac{x - 1}{13} = \frac{y - 2}{-14} = \frac{z - 3}{5} \]

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