If the straight lines $$\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$$ and $$\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$$ intersect at a point, then the integer $$k$$ is equal to

The two given lines are written in symmetric form.

Line 1 : $$\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3}$$
    Point on the line, $$P_1(1,\,2,\,3)$$
    Direction vector, $$\mathbf{a} = \langle k,\,2,\,3\rangle$$
    Parametric form  $$\Rightarrow\; x = 1 + k t,\; y = 2 + 2t,\; z = 3 + 3t\;$$ $$-(1)$$

Line 2 : $$\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2}$$
    Point on the line, $$P_2(2,\,3,\,1)$$
    Direction vector, $$\mathbf{b} = \langle 3,\,k,\,2\rangle$$
    Parametric form  $$\Rightarrow\; x = 2 + 3s,\; y = 3 + ks,\; z = 1 + 2s\;$$ $$-(2)$$

If the lines intersect, there exist real parameters $$t$$ and $$s$$ such that the coordinates from (1) and (2) are equal:

$$1 + k t = 2 + 3s$$ $$-(i)$$
$$2 + 2t = 3 + k s$$ $$-(ii)$$
$$3 + 3t = 1 + 2s$$ $$-(iii)$$

Step 1 → Express $$s$$ in terms of $$t$$
From (iii): $$3 + 3t = 1 + 2s \;\Longrightarrow\; 3t - 2s = -2$$
$$\Rightarrow\; 2s = 3t + 2 \;\Longrightarrow\; s = \frac{3t + 2}{2}$$ $$-(3)$$

Step 2 → Use (i)
Substitute (3) in (i):
$$1 + k t = 2 + 3\left(\frac{3t + 2}{2}\right)$$
$$\Rightarrow\; 1 + k t = 2 + \frac{9t + 6}{2} = \frac{10 + 9t}{2}$$
Multiplying by $$2$$ gives  $$2 + 2kt = 10 + 9t$$
$$\Rightarrow\; t(2k - 9) = 8$$
$$\therefore\; t = \frac{8}{2k - 9}$$ $$-(4)$$

Step 3 → Use (ii)
Substitute (3) in (ii):
$$2 + 2t = 3 + k\left(\frac{3t + 2}{2}\right)$$
Multiplying by $$2$$ gives  $$4 + 4t = 6 + 3k t + 2k$$
Rearrange:
$$t(4 - 3k) - 2(1 + k) = 0$$
$$\therefore\; t = \frac{2(1 + k)}{4 - 3k}$$ $$-(5)$$

Step 4 → Equate the two expressions for $$t$$
From (4) and (5):
$$\frac{8}{2k - 9} = \frac{2(1 + k)}{4 - 3k}$$

Cross-multiply:
$$8(4 - 3k) = 2(1 + k)(2k - 9)$$
$$32 - 24k = 2(1 + k)(2k - 9)$$
Expand the right side:
$$32 - 24k = 2\bigl(2k - 9 + 2k^2 - 9k\bigr)$$
$$32 - 24k = 4k^2 - 14k - 18$$

Collect all terms on one side:
$$0 = 4k^2 - 14k - 18 - 32 + 24k$$
$$0 = 4k^2 + 10k - 50$$
Divide by $$2$$:
$$0 = 2k^2 + 5k - 25$$

Step 5 → Solve the quadratic
Discriminant: $$D = 5^2 - 4(2)(-25) = 25 + 200 = 225 = 15^2$$
$$k = \frac{-5 \pm 15}{4}$$
$$\Rightarrow\; k_1 = \frac{10}{4} = 2.5,\;\; k_2 = \frac{-20}{4} = -5$$

Only integer value: $$k = -5$$.

Denominators in (4) and (5) remain non-zero for $$k = -5$$, so the solution is valid and the lines indeed intersect.

Hence, $$k = -5$$.

Option A which is: $$-5$$