If $a = 7$, $b = 5$, $c = 3$, then the value of $a^2 + b^2 + c^2 - ab - bc - ca$ is

### Concept & Formula The expression $a^2 + b^2 + c^2 - ab - bc - ca$ can be transformed into a sum of squares, which is mathematically elegant and often easier to compute when the values of $a, b,$ and $c$ are in an arithmetic progression (equally spaced). The transformation identity is: $$a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2} \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]$$ ### Step-by-Step Solution * **Given:** $a = 7$, $b = 5$, $c = 3$. * **Calculation (Method 1: Direct Substitution):** $$7^2 + 5^2 + 3^2 - (7)(5) - (5)(3) - (3)(7)$$ $$49 + 25 + 9 - 35 - 15 - 21$$ $$83 - 71 = 12$$ * **Calculation (Method 2: Identity Formula):** Using $\frac{1}{2} \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]$: $$\frac{1}{2} \left[ (7 - 5)^2 + (5 - 3)^2 + (3 - 7)^2 \right]$$ $$\frac{1}{2} \left[ 2^2 + 2^2 + (-4)^2 \right]$$ $$\frac{1}{2} \left[ 4 + 4 + 16 \right] = \frac{1}{2} [24] = 12$$ ### Exam Strategy & Shortcut **Arithmetic Progression Trick:** When $a, b,$ and $c$ form an Arithmetic Progression (AP) with a common difference $d$, the value of $a^2 + b^2 + c^2 - ab - bc - ca$ is always exactly **$3d^2$**. Here, the numbers are $7, 5, 3$. The difference between adjacent terms is $d = 2$. Applying the shortcut: $3 \times (2)^2 = 3 \times 4 = 12$. This solves the question in under 5 seconds! ### Common Pitfall Relying strictly on direct calculation can lead to arithmetic mistakes, especially with larger numbers or negative signs during the subtraction of $ab, bc, ca$. Knowing the identity or the AP shortcut provides a much safer and faster route. ### Final Answer Therefore, the correct answer is **12**.