1. The problem states that point N is the incenter of triangle ABC, and we are given expressions for segments ND and NE. We need to find NF. 2. Since N is the incenter, it is equidistant from all sides of the triangle. This means: $$ND = NE = NF$$ 3. Given: $$ND = 6x - 2$$ $$NE = 3x + 7$$ 4. Set $ND = NE$ to find $x$: $$6x - 2 = 3x + 7$$ 5. Subtract $3x$ from both sides: $$6x - \cancel{3x} - 2 = \cancel{3x} + 7$$ $$3x - 2 = 7$$ 6. Add 2 to both sides: $$3x - 2 + 2 = 7 + 2$$ $$3x = 9$$ 7. Divide both sides by 3: $$\frac{3x}{\cancel{3}} = \frac{9}{\cancel{3}}$$ $$x = 3$$ 8. Substitute $x=3$ into $ND$ to find $NF$: $$NF = ND = 6(3) - 2 = 18 - 2 = 16$$ Final answer: $$\boxed{16}$$