Find the vector equation of line passing through the point `(1,2,-4)` and perpendicular to the two lines: `(x-8)/3=(y+19)/(-16)=(z-10)/7` and `(x-15)/3=(y-29)/8=(z-5)/(-5)`

Options

A. `vecr=(hati+2hatj-4hatk)+lambda(2hati+3hatj+6hatk)`
B. `vecr=(hati+2hatj+4hatk)+lambda(2hati+3hatj+6hatk)`
C. `vecr=(hati-2hatj+4hatk)+lambda(2hati+3hatj+6hatk)`
D. `vecr=(hati+2hatj+4hatk)+lambda(2hati-3hatj-6hatk)`

Answer

To find the vector equation of a line passing through the point \( (1, 2, -4) \) and perpendicular to the two 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:
\[ \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \]
From this, we can extract the direction ratios:
- Direction ratios of the first line: \( \mathbf{d_1} = (3, -16, 7) \)

The second line is given by the equation:
\[ \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} \]
From this, we can extract the direction ratios:
- Direction ratios of the second line: \( \mathbf{d_2} = (3, 8, -5) \)

### Step 2: Find the cross product of the direction ratios
To find a direction vector for the line that is perpendicular to both lines, we need to calculate the cross product of the two direction ratios \( \mathbf{d_1} \) and \( \mathbf{d_2} \).

Let:
\[ \mathbf{d_1} = (3, -16, 7), \quad \mathbf{d_2} = (3, 8, -5) \]

The cross product \( \mathbf{d_1} \times \mathbf{d_2} \) can be calculated using the determinant of the following matrix:
\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} \]

Calculating this determinant:
\[ \mathbf{d_1} \times \mathbf{d_2} = \mathbf{i} \begin{vmatrix} -16 & 7 \\ 8 & -5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 7 \\ 3 & -5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & -16 \\ 3 & 8 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -16 & 7 \\ 8 & -5 \end{vmatrix} = (-16)(-5) - (7)(8) = 80 - 56 = 24 \)
2. \( \begin{vmatrix} 3 & 7 \\ 3 & -5 \end{vmatrix} = (3)(-5) - (7)(3) = -15 - 21 = -36 \)
3. \( \begin{vmatrix} 3 & -16 \\ 3 & 8 \end{vmatrix} = (3)(8) - (-16)(3) = 24 + 48 = 72 \)

Putting it all together:
\[ \mathbf{d_1} \times \mathbf{d_2} = 24 \mathbf{i} + 36 \mathbf{j} + 72 \mathbf{k} \]

### Step 3: Simplify the direction ratios
We can simplify the direction ratios by dividing by 12:
\[ \mathbf{d} = (2, 3, 6) \]

### Step 4: Write the vector equation of the line
The vector equation of a line can be expressed as:
\[ \mathbf{r} = \mathbf{r_0} + \lambda \mathbf{d} \]
where \( \mathbf{r_0} = (1, 2, -4) \) is a point on the line, \( \lambda \) is a parameter, and \( \mathbf{d} = (2, 3, 6) \) is the direction vector.

Thus, the vector equation of the line is:
\[ \mathbf{r} = (1, 2, -4) + \lambda (2, 3, 6) \]

### Final Answer
The vector equation of the required line is:
\[ \mathbf{r} = (1 + 2\lambda, 2 + 3\lambda, -4 + 6\lambda) \]