Step 1: Understanding the Question:
We need to determine the relationship between two given lines in 3D space: whether they intersect or not.
Step 2: Key Formula or Approach:
First, convert the lines into standard symmetrical form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Two lines intersect if they are coplanar and not parallel. Coplanarity condition for lines passing through $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is:
\[ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \]
Step 3: Detailed Explanation:
Line 1: $\frac{6(x-1)}{18} = \frac{y+1}{3} = \frac{z-1}{5} \implies \frac{x-1}{3} = \frac{y+1}{3} = \frac{z-1}{5}$.
Line 1 passes through $A(1, -1, 1)$ with d.r.s $(3, 3, 5)$.
Line 2: $\frac{3(x+2)}{12} = \frac{y-1}{3} = \frac{z+1}{2} \implies \frac{x+2}{4} = \frac{y-1}{3} = \frac{z+1}{2}$.
Line 2 passes through $B(-2, 1, -1)$ with d.r.s $(4, 3, 2)$.
Let's check the determinant for coplanarity:
\[ D = \begin{vmatrix} -2-1 & 1-(-1) & -1-1\\ 3 & 3 & 5 \\ 4 & 3 & 2 \end{vmatrix} = \begin{vmatrix} -3 & 2 & -2\\ 3 & 3 & 5 \\ 4 & 3 & 2 \end{vmatrix} \]
Expanding along the first row:
\[ D = -3(6 - 15) - 2(6 - 20) - 2(9 - 12) \]
\[ D = -3(-9) - 2(-14) - 2(-3) = 27 + 28 + 6 = 61 \]
Since $D \neq 0$, the lines are non-coplanar (skew lines) and thus do not intersect.
Step 4: Final Answer:
The lines do not intersect.