Question:

Step 1: Apply the Law of Cosines to express $\cos A$ in terms of the sides. By the Law of Cosines, we have: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ Dividing both sides by $a$: $$\frac{\cos A}{a} = \frac{b^2 + c^2 - a^2}{2abc}$$ Step 2: Define the common value and express $k_1$ in terms of it. Let $S$ denote the common value: $$S = \frac{a^2 + b^2 + c^2}{8}$$ Since $\frac{\cos A}{a} + k_1 = S$, we can write: $$k_1 = S - \frac{\cos A}{a} = \frac{a^2 + b^2 + c^2}{8} - \frac{b^2 + c^2 - a^2}{2abc}$$ Step 3: Simplify $k_1$ using the condition $abc = 4$. Since $abc = 4$, we have $2abc = 8$. Substituting this: $$k_1 = \frac{a^2 + b^2 + c^2}{8} - \frac{b^2 + c^2 - a^2}{8}$$ Combining the fractions: $$k_1 = \frac{a^2 + b^2 + c^2 - (b^2 + c^2 - a^2)}{8} = \frac{a^2 + b^2 + c^2 - b^2 - c^2 + a^2}{8} = \frac{2a^2}{8} = \frac{a^2}{4}$$ Step 4: Find $k_2$ and $k_3$ using the same approach. By applying the same reasoning to angles $B$ and $C$: $$k_2 = \frac{b^2}{4}$$ $$k_3 = \frac{c^2}{4}$$ Step 5: Calculate the product $k_1 k_2 k_3$. $$k_1 k_2 k_3 = \frac{a^2}{4} \cdot \frac{b^2}{4} \cdot \frac{c^2}{4} = \frac{a^2 b^2 c^2}{64} = \frac{(abc)^2}{64}$$ Step 6: Substitute $abc = 4$ and find the final answer. $$k_1 k_2 k_3 = \frac{(4)^2}{64} = \frac{16}{64} = \frac{1}{4}$$ Therefore, the value of $k_1 k_2 k_3$ is $\boxed{\frac{1}{4}}$, which corresponds to **Option 4**.