1. The problem involves subtracting two scalar multiples of vectors: $$\frac{3}{7}(2, 8a - 2, 1) - \frac{1}{9}(2, 7a - 3, 6)$$ 2. Multiply each vector by its scalar: $$\frac{3}{7} \times (2, 8a - 2, 1) = \left(\frac{6}{7}, \frac{24a - 6}{7}, \frac{3}{7}\right)$$ $$\frac{1}{9} \times (2, 7a - 3, 6) = \left(\frac{2}{9}, \frac{7a - 3}{9}, \frac{6}{9}\right)$$ 3. Subtract the vectors component-wise: $$\left(\frac{6}{7} - \frac{2}{9}, \frac{24a - 6}{7} - \frac{7a - 3}{9}, \frac{3}{7} - \frac{6}{9}\right)$$ 4. Simplify each component: - For the first component: $$\frac{6}{7} - \frac{2}{9} = \frac{6 \times 9}{7 \times 9} - \frac{2 \times 7}{9 \times 7} = \frac{54}{63} - \frac{14}{63} = \frac{40}{63}$$ - For the second component: $$\frac{24a - 6}{7} - \frac{7a - 3}{9} = \frac{(24a - 6) \times 9}{63} - \frac{(7a - 3) \times 7}{63} = \frac{216a - 54 - 49a + 21}{63} = \frac{167a - 33}{63}$$ - For the third component: $$\frac{3}{7} - \frac{6}{9} = \frac{3}{7} - \frac{2}{3} = \frac{3 \times 3}{21} - \frac{2 \times 7}{21} = \frac{9}{21} - \frac{14}{21} = -\frac{5}{21}$$ 5. Therefore, the final vector is: $$\left( \frac{40}{63}, \frac{167a - 33}{63}, -\frac{5}{21} \right)$$