Question:

Step 1: Apply the Remainder Theorem to find values of the polynomial. When a polynomial $P(x)$ is divided by $(x-1)$, $(x-2)$, and $(x-3)$ with remainders 4, 5, and 6 respectively, by the Remainder Theorem: $$P(1) = 4, \quad P(2) = 5, \quad P(3) = 6$$ Step 2: Set up equations using the division algorithm. When $P(x)$ is divided by $(x-1)(x-2)(x-3)$, we can write: $$P(x) = Q(x) \cdot (x-1)(x-2)(x-3) + R(x)$$ where $R(x) = ax^2 + bx + c$ is the remainder (a polynomial of degree less than 3). Since this equation holds for all $x$, it must hold at $x = 1, 2, 3$: $$P(1) = R(1) \implies a + b + c = 4 \quad \text{...(1)}$$ $$P(2) = R(2) \implies 4a + 2b + c = 5 \quad \text{...(2)}$$ $$P(3) = R(3) \implies 9a + 3b + c = 6 \quad \text{...(3)}$$ Step 3: Solve the system of linear equations. Subtract equation (1) from equation (2): $$(4a + 2b + c) - (a + b + c) = 5 - 4$$ $$3a + b = 1 \quad \text{...(4)}$$ Subtract equation (2) from equation (3): $$(9a + 3b + c) - (4a + 2b + c) = 6 - 5$$ $$5a + b = 1 \quad \text{...(5)}$$ Subtract equation (4) from equation (5): $$(5a + b) - (3a + b) = 1 - 1$$ $$2a = 0 \implies a = 0$$ Substitute $a = 0$ into equation (4): $$3(0) + b = 1 \implies b = 1$$ Substitute $a = 0$ and $b = 1$ into equation (1): $$0 + 1 + c = 4 \implies c = 3$$ Step 4: Calculate the final answer. Now we compute $3a + 2b + c$: $$3a + 2b + c = 3(0) + 2(1) + 3 = 0 + 2 + 3 = 5$$ The answer is **5**, which corresponds to **Option 3**.