The lines with vector equations are, `vecr_(1)=3hati+6hatj+lambda(-4hati+3hatj+2hatk) and vecr_(2)=-2hati+7hatj+mu(-4hati+hatj+hatk)` are such that :

Options

A. they are coplanar
B. they do not intersect
C. they are skew
D. the angle between then is `tan^(-1)(3//7)`

Answer

To solve the problem, we need to analyze the given vector equations of the two lines and determine their relationship. The vector equations are:

1. \( \vec{r_1} = 3\hat{i} + 6\hat{j} + \lambda(-4\hat{i} + 3\hat{j} + 2\hat{k}) \)
2. \( \vec{r_2} = -2\hat{i} + 7\hat{j} + \mu(-4\hat{i} + \hat{j} + \hat{k}) \)

### Step 1: Identify Position and Direction Vectors
From the equations, we can identify the position and direction vectors for both lines.

- For line \( \vec{r_1} \):
- Position vector \( \vec{l_1} = 3\hat{i} + 6\hat{j} \)
- Direction vector \( \vec{m_1} = -4\hat{i} + 3\hat{j} + 2\hat{k} \)

- For line \( \vec{r_2} \):
- Position vector \( \vec{l_2} = -2\hat{i} + 7\hat{j} \)
- Direction vector \( \vec{m_2} = -4\hat{i} + \hat{j} + \hat{k} \)

### Step 2: Check for Coplanarity
To check if the lines are coplanar, we use the condition:
\[ (\vec{l_2} - \vec{l_1}) \cdot (\vec{m_1} \times \vec{m_2}) = 0 \]

First, calculate \( \vec{l_2} - \vec{l_1} \):
\[ \vec{l_2} - \vec{l_1} = (-2\hat{i} + 7\hat{j}) - (3\hat{i} + 6\hat{j}) = -5\hat{i} + \hat{j} \]

Next, calculate \( \vec{m_1} \times \vec{m_2} \):
\[ \vec{m_1} = \begin{pmatrix} -4 \\ 3 \\ 2 \end{pmatrix}, \quad \vec{m_2} = \begin{pmatrix} -4 \\ 1 \\ 1 \end{pmatrix} \]

Using the determinant method:
\[ \vec{m_1} \times \vec{m_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 3 & 2 \\ -4 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(3 \cdot 1 - 2 \cdot 1) - \hat{j}(-4 \cdot 1 - 2 \cdot -4) + \hat{k}(-4 \cdot 1 - 3 \cdot -4) \]
\[ = \hat{i}(3 - 2) - \hat{j}(-4 + 8) + \hat{k}(-4 + 12) \]
\[ = \hat{i}(1) - \hat{j}(4) + \hat{k}(8) = \hat{i} - 4\hat{j} + 8\hat{k} \]

Now, compute the dot product:
\[ (-5\hat{i} + \hat{j}) \cdot (\hat{i} - 4\hat{j} + 8\hat{k}) = (-5)(1) + (1)(-4) + (0)(8) = -5 - 4 = -9 \]

Since the dot product is not equal to zero, the lines are **not coplanar**.

### Step 3: Check for Intersection
For the lines to intersect, we need to set \( \vec{r_1} = \vec{r_2} \).

Equating the components:
1. \( 3 - 4\lambda = -2 - 4\mu \)
2. \( 6 + 3\lambda = 7 + \mu \)
3. \( 2\lambda = \mu \)

From the third equation, substitute \( \mu = 2\lambda \) into the first two equations:
1. \( 3 - 4\lambda = -2 - 4(2\lambda) \)
\[ 3 - 4\lambda = -2 - 8\lambda \implies 4\lambda - 8\lambda = -2 - 3 \implies -4\lambda = -5 \implies \lambda = \frac{5}{4} \]

2. Substitute \( \lambda \) into \( \mu = 2\lambda \):
\[ \mu = 2 \cdot \frac{5}{4} = \frac{5}{2} \]

Now, substitute \( \lambda \) and \( \mu \) back into the first equation:
\[ 3 - 4\left(\frac{5}{4}\right) = -2 - 4\left(\frac{5}{2}\right) \]
Calculating both sides:
\[ 3 - 5 = -2 - 10 \implies -2 = -12 \text{ (not equal)} \]

Since there are no values of \( \lambda \) and \( \mu \) that satisfy all equations, the lines **do not intersect**.

### Step 4: Check for Skewness
Since the lines are not parallel (as shown by the non-zero cross product) and do not intersect, they are classified as **skew lines**.

### Step 5: Calculate the Angle Between the Lines
The angle \( \theta \) between the two lines can be found using the formula:
\[ \cos \theta = \frac{\vec{m_1} \cdot \vec{m_2}}{|\vec{m_1}| |\vec{m_2}|} \]

Calculate \( \vec{m_1} \cdot \vec{m_2} \):
\[ \vec{m_1} \cdot \vec{m_2} = (-4)(-4) + (3)(1) + (2)(1) = 16 + 3 + 2 = 21 \]

Now calculate the magnitudes:
\[ |\vec{m_1}| = \sqrt{(-4)^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29} \]
\[ |\vec{m_2}| = \sqrt{(-4)^2 + 1^2 + 1^2} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2} \]

Now substitute into the cosine formula:
\[ \cos \theta = \frac{21}{\sqrt{29} \cdot 3\sqrt{2}} = \frac{21}{3\sqrt{58}} = \frac{7}{\sqrt{58}} \]

### Conclusion
The lines are:
- Not coplanar
- Do not intersect
- Skew lines
- The angle between them is given by \( \theta = \cos^{-1}\left(\frac{7}{\sqrt{58}}\right) \)