If the lines $$\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k}$$ and $$\frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1}$$ are coplanar, then $$k$$ can have

We have the first line in its symmetric (Cartesian) form

$$\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$$

From this, the direction ratios are read directly as the denominators of the three fractions. Hence the direction vector of the first line is

$$\vec{a}= \langle 1,\,1,\,-k\rangle$$

and a convenient point on the line is

$$P_1(2,\,3,\,4).$$

Similarly, the second line is

$$\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1},$$

so its direction vector is

$$\vec{b}=\langle k,\,2,\,1\rangle$$

and a point on it is

$$P_2(1,\,4,\,5).$$

To decide whether the two lines are coplanar (i.e. not skew), we recall the standard condition:

For two lines with direction vectors $$\vec{a}$$ and $$\vec{b}$$ and position vectors of points $$\vec{r}_1$$ and $$\vec{r}_2$$ on them, the lines are coplanar iff

$$(\vec{a}\times\vec{b})\cdot(\vec{r}_2-\vec{r}_1)=0.$$

First we calculate $$\vec{r}_2-\vec{r}_1,$$ the vector joining the given points:

$$\vec{r}_2-\vec{r}_1=P_2P_1=\langle 1-2,\; 4-3,\; 5-4\rangle =\langle -1,\;1,\;1\rangle.$$

Next we compute the cross product $$\vec{a}\times\vec{b}.$$ Using the determinant form,

$$ \vec{a}\times\vec{b} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&1&-k\\ k&2&1 \end{vmatrix}. $$

Expanding this determinant, we get

$$ \vec{a}\times\vec{b} = \mathbf{i}\bigl(1\cdot1-(-k)\cdot2\bigr) -\mathbf{j}\bigl(1\cdot1-(-k)\cdot k\bigr) +\mathbf{k}\bigl(1\cdot2-1\cdot k\bigr). $$

Simplifying term by term,

$$ \vec{a}\times\vec{b} = \mathbf{i}(1+2k)\; -\mathbf{j}(1+k^{2})\; +\mathbf{k}(2-k). $$

In coordinate form this is

$$ \vec{a}\times\vec{b}= \langle 1+2k,\; -(1+k^2),\; 2-k\rangle. $$

Now we take the scalar triple product with $$\vec{r}_2-\vec{r}_1=\langle -1,\;1,\;1\rangle$$ and set it equal to zero:

$$ (\vec{a}\times\vec{b})\cdot(\vec{r}_2-\vec{r}_1)= \langle 1+2k,\; -(1+k^2),\; 2-k\rangle\cdot\langle -1,\;1,\;1\rangle=0. $$

Carrying out the dot product component-wise,

$$ (1+2k)(-1) + \bigl[-(1+k^2)\bigr](1) + (2-k)(1)=0. $$

Let us open every bracket carefully:

$$ -(1+2k) - (1+k^2) + (2-k)=0. $$

Now collect like terms. First the constant terms:

$$ -1 -1 + 2 = 0. $$

Next the terms in $$k$$:

$$ -2k - k = -3k. $$

Finally the term in $$k^2$$:

$$ -\,k^2. $$

Thus the entire expression reduces to

$$ -\,k^2 - 3k = 0. $$

We can remove the overall minus sign (it does not affect the equality) to obtain

$$ k^2 + 3k = 0. $$

Factoring out $$k,$$

$$ k(k+3)=0. $$

This gives the two possible values

$$ k = 0 \quad \text{or} \quad k = -3. $$

No other value satisfies the coplanarity condition, so exactly two real values of $$k$$ are possible.

Hence, the correct answer is Option A.