1. Write the two lines in parametric form

For the first line, $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}=\lambda$ So, $x=1+k\lambda,\quad y=2+2\lambda,\quad z=3+3\lambda.$

For the second line, $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}=\mu$ So, $x=2+3\mu,\quad y=3+k\mu,\quad z=1+2\mu.$

2. Since the lines intersect, there exist $\lambda,\mu$ such that the coordinates are equal

Thus, $1+k\lambda=2+3\mu \quad ...(1)$ $2+2\lambda=3+k\mu \quad ...(2)$ $3+3\lambda=1+2\mu \quad ...(3)$

3. First solve for $\lambda$ and $\mu$ using equations (1) and (3)

From (3): $3\lambda-2\mu=-2$

From (1): $k\lambda-3\mu=1$

But it is easier to first use (3) to express one variable.

From (3), $3\lambda-2\mu=-2.$

Now solve using the actual coordinate equations more directly.

From (3): $3+3\lambda=1+2\mu$ $3\lambda-2\mu=-2. \quad ...(3')$

Let us solve (1) and (3').

From (1): $k\lambda-3\mu=1. \quad ...(1')$

From (3'): $\mu=\frac{3\lambda+2}{2}.$

Substitute into (1'): $k\lambda-3\left(\frac{3\lambda+2}{2}\right)=1$ $2k\lambda-9\lambda-6=2$ $(2k-9)\lambda=8$ $\lambda=\frac{8}{2k-9}.$

Then $\mu=\frac{3\lambda+2}{2}=\frac{3\cdot \frac{8}{2k-9}+2}{2}.$

Instead of continuing this way, a cleaner approach is to eliminate using equations (2) and (3).

4. Use equations (2) and (3) to get $\lambda,\mu$

From (3): $3+3\lambda=1+2\mu$ $3\lambda-2\mu=-2. \quad ...(A)$

From (2): $2+2\lambda=3+k\mu$ $2\lambda-k\mu=1. \quad ...(B)$

Solve (A) and (B).

Multiply (B) by $3$: $6\lambda-3k\mu=3. \quad ...(C)$

Multiply (A) by $2$: $6\lambda-4\mu=-4. \quad ...(D)$

Subtract (D) from (C): $(-3k+4)\mu=7$ $\mu=\frac{7}{4-3k}.$

Then from (A): $3\lambda= -2+2\mu$ $\lambda=\frac{-2+2\mu}{3}.$

Now substitute into equation (1): $1+k\lambda=2+3\mu$ $k\lambda-3\mu=1.$

Substitute $\lambda=\frac{-2+2\mu}{3}$: $k\left(\frac{-2+2\mu}{3}\right)-3\mu=1$ $-2k+2k\mu-9\mu=3$ $\mu(2k-9)=2k+3$ $\mu=\frac{2k+3}{2k-9}.$

But we also have $\mu=\frac{7}{4-3k}.$

Hence, $\frac{2k+3}{2k-9}=\frac{7}{4-3k}.$

Cross-multiplying: $(2k+3)(4-3k)=7(2k-9).$

Expand: $8k+12-6k^2-9k=14k-63$ $-6k^2-k+12=14k-63$ $-6k^2-15k+75=0$ $6k^2+15k-75=0$ $2k^2+5k-25=0.$

Factor: $2k^2+10k-5k-25=0$ $2k(k+5)-5(k+5)=0$ $(2k-5)(k+5)=0.$

So, $k=\frac{5}{2} \quad \text{or} \quad k=-5.$

5. Choose the integer value

Since the question asks for the integer $k$, among the options only $k=-5$ is valid.

6. Compare with stored answer

Stored correct answer: A = $-5$.

Our derived answer is also $-5$, so it agrees.