If the lines $$\frac{x+1}{2} = \frac{y-1}{1} = \frac{z+1}{3}$$ and $$\frac{x+2}{2} = \frac{y-k}{3} = \frac{z}{4}$$ are coplanar, then the value of $$k$$ is :

To determine the value of $$k$$ for which the given lines are coplanar, we start by recalling that two lines are coplanar if they lie in the same plane. The condition for coplanarity can be expressed using the scalar triple product. Specifically, if we have two lines:

Line 1: Passes through point $$A$$ with position vector $$\vec{a}$$ and direction vector $$\vec{b}$$.

Line 2: Passes through point $$B$$ with position vector $$\vec{c}$$ and direction vector $$\vec{d}$$.

The lines are coplanar if the vector $$\overrightarrow{AB} = \vec{c} - \vec{a}$$ is perpendicular to the cross product $$\vec{b} \times \vec{d}$$. This gives the condition:

$$$ (\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d}) = 0 $$$

Now, let's identify the points and direction vectors for the given lines.

The first line is $$\frac{x+1}{2} = \frac{y-1}{1} = \frac{z+1}{3}$$. This can be rewritten as $$\frac{x - (-1)}{2} = \frac{y - 1}{1} = \frac{z - (-1)}{3}$$. So, a point on this line is $$A(-1, 1, -1)$$, and the direction vector $$\vec{b} = (2, 1, 3)$$.

The second line is $$\frac{x+2}{2} = \frac{y-k}{3} = \frac{z}{4}$$. This can be rewritten as $$\frac{x - (-2)}{2} = \frac{y - k}{3} = \frac{z - 0}{4}$$. So, a point on this line is $$B(-2, k, 0)$$, and the direction vector $$\vec{d} = (2, 3, 4)$$.

The vector $$\overrightarrow{AB}$$ is:

$$$ \overrightarrow{AB} = (-2 - (-1), k - 1, 0 - (-1)) = (-1, k-1, 1) $$$

Next, we compute the cross product $$\vec{b} \times \vec{d}$$. Given $$\vec{b} = (2, 1, 3)$$ and $$\vec{d} = (2, 3, 4)$$, the cross product is:

$$$ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 2 & 3 & 4 \\ \end{vmatrix} $$$

Expanding the determinant:

$$$ \hat{i}(1 \cdot 4 - 3 \cdot 3) - \hat{j}(2 \cdot 4 - 3 \cdot 2) + \hat{k}(2 \cdot 3 - 1 \cdot 2) $$$

Calculating each component:

$$$ \hat{i}(4 - 9) = \hat{i}(-5) $$$

$$$ -\hat{j}(8 - 6) = -\hat{j}(2) = -2\hat{j} $$$

$$$ \hat{k}(6 - 2) = \hat{k}(4) $$$

So, $$\vec{b} \times \vec{d} = (-5, -2, 4)$$.

Now, the scalar triple product condition is:

$$$ \overrightarrow{AB} \cdot (\vec{b} \times \vec{d}) = (-1, k-1, 1) \cdot (-5, -2, 4) = 0 $$$

Computing the dot product:

$$$ (-1) \cdot (-5) + (k-1) \cdot (-2) + (1) \cdot (4) = 0 $$$

Simplifying:

$$$ 5 - 2(k-1) + 4 = 0 $$$

Combine the constant terms:

$$$ 5 + 4 = 9 $$$

So:

$$$ 9 - 2(k-1) = 0 $$$

Distribute the $$-2$$:

$$$ 9 - 2k + 2 = 0 $$$

Combine constants:

$$$ 11 - 2k = 0 $$$

Solve for $$k$$:

$$$ -2k = -11 $$$

$$$ k = \frac{-11}{-2} = \frac{11}{2} $$$

Thus, the value of $$k$$ is $$\frac{11}{2}$$. Comparing with the options, this corresponds to Option A.

Hence, the correct answer is Option A.