1. **State the problem:** We have a linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by its action on the standard basis vectors: $$T(1,0,0) = (2,2,3), \quad T(0,1,0) = (4,5,1), \quad T(0,0,1) = (3,1,7)$$ We want to find $T(2,4,1)$. 2. **Recall the linearity property:** For any vectors $\mathbf{u}, \mathbf{v}$ and scalars $a, b$, a linear transformation satisfies: $$T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v})$$ 3. **Express the vector $(2,4,1)$ as a linear combination of the basis vectors:** $$(2,4,1) = 2(1,0,0) + 4(0,1,0) + 1(0,0,1)$$ 4. **Apply $T$ using linearity:** $$T(2,4,1) = 2T(1,0,0) + 4T(0,1,0) + 1T(0,0,1)$$ 5. **Substitute the given values:** $$= 2(2,2,3) + 4(4,5,1) + 1(3,1,7)$$ 6. **Calculate each term:** $$2(2,2,3) = (4,4,6)$$ $$4(4,5,1) = (16,20,4)$$ $$1(3,1,7) = (3,1,7)$$ 7. **Add the vectors component-wise:** $$ (4,4,6) + (16,20,4) + (3,1,7) = (4+16+3, 4+20+1, 6+4+7) = (23, 25, 17)$$ **Final answer:** $$\boxed{T(2,4,1) = (23, 25, 17)}$$