The shortest distance between lines $\frac{-x-3}{4}=\frac{y-6}{3}=\frac{z}{2}$ and $\frac{-x-2}{4}=\frac{y}{1}=\frac{z-7}{1}$ is:

9 units

The correct answer is Option (3) → 9 units

Given lines:

\(\frac{-x - 3}{4} = \frac{y - 6}{3} = \frac{z}{2}\)

\(\frac{-x - 2}{4} = \frac{y}{1} = \frac{z - 7}{1}\)

Parametric form:

Line 1:

\[ \begin{cases} \frac{-x - 3}{4} = t \Rightarrow x = -4t - 3 \\ \frac{y - 6}{3} = t \Rightarrow y = 3t + 6 \\ \frac{z}{2} = t \Rightarrow z = 2t \end{cases} \]

Line 2:

\[ \begin{cases} \frac{-x - 2}{4} = s \Rightarrow x = -4s - 2 \\ \frac{y}{1} = s \Rightarrow y = s \\ \frac{z - 7}{1} = s \Rightarrow z = s + 7 \end{cases} \]

Direction vectors:

\(\vec{d_1} = (-4, 3, 2)\)

\(\vec{d_2} = (-4, 1, 1)\)

Points on lines at parameter 0:

\(\vec{A} = (-3, 6, 0)\)

\(\vec{B} = (-2, 0, 7)\)

Vector between points:

\(\vec{AB} = \vec{B} - \vec{A} = (1, -6, 7)\)

Cross product:

\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 2 \\ -4 & 1 & 1 \\ \end{vmatrix} = (1, -4, 8) \]

Magnitude of cross product:

\(|\vec{d_1} \times \vec{d_2}| = \sqrt{1^2 + (-4)^2 + 8^2} = 9\)

Dot product:

\(\vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 81\)

Shortest distance:

\[ \frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} = \frac{81}{9} = 9 \]