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 regarding the coplanarity and angle between the given lines with vector equations, we will follow these steps:

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

### Step 1: Identify the direction vectors and points on the lines
- For Line 1, the point \(\vec{A} = 3\hat{i} + 6\hat{j}\) and the direction vector \(\vec{b_1} = -4\hat{i} + 3\hat{j} + 2\hat{k}\).
- For Line 2, the point \(\vec{B} = -2\hat{i} + 7\hat{j}\) and the direction vector \(\vec{b_2} = -4\hat{i} + \hat{j} + \hat{k}\).

### Step 2: Check for coplanarity
To check if the lines are coplanar, we can use the scalar triple product. The lines are coplanar if:
\[ (\vec{B} - \vec{A}) \cdot (\vec{b_1} \times \vec{b_2}) = 0 \]

Calculate \(\vec{B} - \vec{A}\):
\[ \vec{B} - \vec{A} = (-2 - 3)\hat{i} + (7 - 6)\hat{j} + (0 - 0)\hat{k} = -5\hat{i} + 1\hat{j} + 0\hat{k} \]

Now, compute \(\vec{b_1} \times \vec{b_2}\):
\[ \vec{b_1} = -4\hat{i} + 3\hat{j} + 2\hat{k}, \quad \vec{b_2} = -4\hat{i} + 1\hat{j} + 1\hat{k} \]

Using the determinant to find the cross product:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 3 & 2 \\ -4 & 1 & 1 \end{vmatrix} \]

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:
\[ (\vec{B} - \vec{A}) \cdot (\vec{b_1} \times \vec{b_2}) = (-5\hat{i} + 1\hat{j} + 0\hat{k}) \cdot (1\hat{i} - 4\hat{j} + 8\hat{k}) = -5 \cdot 1 + 1 \cdot -4 + 0 \cdot 8 = -5 - 4 + 0 = -9 \]

Since the result is not zero, the lines are **not coplanar**.

### Step 3: Determine if the lines are skew
Since the lines are not coplanar and do not intersect, they are skew lines.

### Step 4: Find the angle between the lines
The angle \(\theta\) between the two lines can be found using the formula:
\[ \cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|} \]

Calculating \(\vec{b_1} \cdot \vec{b_2}\):
\[ \vec{b_1} \cdot \vec{b_2} = (-4)(-4) + (3)(1) + (2)(1) = 16 + 3 + 2 = 21 \]

Calculating the magnitudes:
\[ |\vec{b_1}| = \sqrt{(-4)^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29} \]
\[ |\vec{b_2}| = \sqrt{(-4)^2 + 1^2 + 1^2} = \sqrt{16 + 1 + 1} = \sqrt{18} \]

Now substituting into the cosine formula:
\[ \cos \theta = \frac{21}{\sqrt{29} \cdot \sqrt{18}} = \frac{21}{\sqrt{522}} = \frac{21}{3\sqrt{58}} = \frac{7}{\sqrt{58}} \]

### Step 5: Find \(\tan \theta\)
Using the relationship between sine and cosine:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Let \(\cos \theta = \frac{7}{\sqrt{58}}\), then:
\[ \sin^2 \theta = 1 - \left(\frac{7}{\sqrt{58}}\right)^2 = 1 - \frac{49}{58} = \frac{9}{58} \]
\[ \sin \theta = \frac{3}{\sqrt{58}} \]

Now, \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{\sqrt{58}}}{\frac{7}{\sqrt{58}}} = \frac{3}{7}\)

Thus, the angle between the lines is:
\[ \theta = \tan^{-1}\left(\frac{3}{7}\right) \]

### Conclusion
1. The lines are **not coplanar** and are **skew**.
2. The angle between the lines is \(\tan^{-1}\left(\frac{3}{7}\right)\).