`L_(1):(x+1)/(-3)=(y-3)/(2)=(z+2)/(1),L_(2):(x)/(1)=(y-7)/(-3)=(z+7)/(2)`
The lines `L_(1)` and `L_(2)` are -

Options

A. Perpendicular
B. Parallel
C. Coplanar
D. None of these

Answer

To determine the relationship between the lines \( L_1 \) and \( L_2 \), we will analyze their direction ratios and check for coplanarity.

### Step 1: Identify the direction ratios of the lines

The equations of the lines are given in symmetric form:

1. For line \( L_1 \):
\[ \frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1} \]
The direction ratios \( B_1 \) for line \( L_1 \) are:
\[ B_1 = (-3, 2, 1) \]

2. For line \( L_2 \):
\[ \frac{x}{1} = \frac{y - 7}{-3} = \frac{z + 7}{2} \]
The direction ratios \( B_2 \) for line \( L_2 \) are:
\[ B_2 = (1, -3, 2) \]

### Step 2: Check if the lines are perpendicular

To check if the lines are perpendicular, we calculate the dot product of their direction ratios:
\[ B_1 \cdot B_2 = (-3)(1) + (2)(-3) + (1)(2) \]
Calculating this gives:
\[ -3 - 6 + 2 = -7 \]
Since the dot product is not zero, the lines are **not perpendicular**.

### Step 3: Check if the lines are parallel

For the lines to be parallel, their direction ratios must be proportional. We check if there exists a scalar \( k \) such that:
\[ B_1 = k B_2 \]
This means:
\[ (-3, 2, 1) = k(1, -3, 2) \]
From the first component:
\[ -3 = k \cdot 1 \implies k = -3 \]
From the second component:
\[ 2 = -3 \cdot (-3) \implies 2 = 9 \quad \text{(not true)} \]
Since the ratios are not consistent, the lines are **not parallel**.

### Step 4: Check if the lines are coplanar

To check for coplanarity, we use the determinant condition. The lines are coplanar if the following determinant is zero:
\[ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \]

Where:
- \( (x_1, y_1, z_1) \) are the coordinates from line \( L_1 \) (taking \( x = -1, y = 3, z = -2 \))
- \( (x_2, y_2, z_2) \) are the coordinates from line \( L_2 \) (taking \( x = 0, y = 7, z = -7 \))
- \( (a_1, b_1, c_1) = (-3, 2, 1) \)
- \( (a_2, b_2, c_2) = (1, -3, 2) \)

Calculating the differences:
\[ x_2 - x_1 = 0 - (-1) = 1 \]
\[ y_2 - y_1 = 7 - 3 = 4 \]
\[ z_2 - z_1 = -7 - (-2) = -5 \]

Now substituting into the determinant:
\[ \begin{vmatrix} 1 & 4 & -5 \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = 1 \begin{vmatrix} 2 & 1 \\ -3 & 2 \end{vmatrix} - 4 \begin{vmatrix} -3 & 1 \\ 1 & 2 \end{vmatrix} - 5 \begin{vmatrix} -3 & 2 \\ 1 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 1 \\ -3 & 2 \end{vmatrix} = (2)(2) - (1)(-3) = 4 + 3 = 7 \)
2. \( \begin{vmatrix} -3 & 1 \\ 1 & 2 \end{vmatrix} = (-3)(2) - (1)(1) = -6 - 1 = -7 \)
3. \( \begin{vmatrix} -3 & 2 \\ 1 & -3 \end{vmatrix} = (-3)(-3) - (2)(1) = 9 - 2 = 7 \)

Now substituting back:
\[ = 1(7) - 4(-7) - 5(7) = 7 + 28 - 35 = 0 \]

Since the determinant is zero, the lines are **coplanar**.

### Conclusion

The lines \( L_1 \) and \( L_2 \) are **coplanar**.

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