1. **State the problem:** Find the angle between the two lines given by the symmetric equations: $$\frac{x}{2} = -(y+1) = \frac{z+1}{-2}$$ and $$x - 6 = y = \frac{1}{2}(z-1)$$ 2. **Extract direction vectors:** For the first line, set the common ratio to parameter $t$: $$\frac{x}{2} = t \Rightarrow x = 2t$$ $$-(y+1) = t \Rightarrow y = -1 - t$$ $$\frac{z+1}{-2} = t \Rightarrow z = -1 - 2t$$ Direction vector of line 1 is: $$\vec{d_1} = \langle 2, -1, -2 \rangle$$ For the second line, set the common ratio to parameter $s$: $$x - 6 = s \Rightarrow x = 6 + s$$ $$y = s$$ $$\frac{1}{2}(z - 1) = s \Rightarrow z = 1 + 2s$$ Direction vector of line 2 is: $$\vec{d_2} = \langle 1, 1, 2 \rangle$$ 3. **Formula for angle between two lines:** The angle $\theta$ between two lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$ is given by: $$\cos \theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{||\vec{d_1}|| \cdot ||\vec{d_2}||}$$ where $\cdot$ denotes the dot product and $||\cdot||$ denotes the vector magnitude. 4. **Calculate dot product:** $$\vec{d_1} \cdot \vec{d_2} = (2)(1) + (-1)(1) + (-2)(2) = 2 - 1 - 4 = -3$$ Take absolute value: $$|\vec{d_1} \cdot \vec{d_2}| = 3$$ 5. **Calculate magnitudes:** $$||\vec{d_1}|| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ $$||\vec{d_2}|| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$ 6. **Calculate cosine of angle:** $$\cos \theta = \frac{3}{3 \times \sqrt{6}} = \frac{1}{\sqrt{6}}$$ 7. **Calculate angle $\theta$:** $$\theta = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 65.91^\circ$$ 8. **Interpretation:** The angle between the lines is approximately $65.91^\circ$. The options given are $45^\circ$, $47.2^\circ$, $30^\circ$, and $90^\circ$. None match exactly, but the closest is none. Possibly a mistake in options or problem statement. **Final answer:** Approximately $65.91^\circ$ (not among given options).