Since $Q_{1}, Q{2}, Q_{3}$ are orthogonal

$\int\limits_{a}^{b}[Q_{1}(x)]^2 dx = \int\limits_{a}^{b}[Q_{2}(x)]^2dx = \int\limits_{a}^{b}[Q_{3}(x)]^2 dx = 1$.....(1)

And $\int\limits_{a}^{b}Q_{m}(x) Q_{n}(x)dx = 0$ where $m \neq n$....(2)

Now if $\int\limits_{a}^{b}[f(x)]^2dx = \int\limits_{a}^{b}[C_{1}Q_{1}(x) + C_{2}Q_{2}(x) + C_{3}Q_{3}(x)]^2 dx$

$\int\limits_{a}^{b}[f(x)]^2dx = \int\limits_{a}^{b}[C_{1}Q_{1}(x) + C_{2}Q_{2}(x) + C_{3}Q_{3}(x)] * [C_{1}Q_{1}x + C_{2}Q_{2}(x) + C_{3}Q_{3}(x)]dx$

$\int\limits_{a}^{b}[f(x)]^2dx = \int\limits_{a}^{b}[C_{1}^2 Q_{1}(x)^2 + C_{1}C_{2}Q_{1}(x)Q_{2}(x) + C_{1}C_{3}Q_{1}(x)Q_{2}(x) + C_{1}C_{2}Q_{1}(x)Q_{2}(x) + C_{2}^2 Q_{2}(x)^2 + C_{2}C_{3}Q_{2}(x) Q_{3}(x) + C_{1}C_{3}Q_{1}(x)Q_{3}(x) + C_{3}C_{2}Q_{2}(x)Q_{3}(x) + C_{3}^2 Q_{3}(x)^2]dx$

$\int\limits_{a}^{b}[f(x)]^2dx = \int\limits_{a}^{b}[C_{1}^2 Q_{1}(x)^2 + C_{2}^2 Q_{2}(x)^2 + C_{3}^2 Q_{3}(x)^2 + 2C_{1} C_{2} Q_{1}(x) Q_{2}(x) + 2C_{2}C_{3}Q_{2}(x)Q_{3}(x) + 2C_{1}C_{3}Q_{1}(x)Q_{3}(x)]dx$......(3)

From (1) & (2) equation (3) can be written as

$\int\limits_{a}^{b}[f(x)]^2dx = C_{1}^2 + C_{2}^2 + C_{3}^2$

Since all the square terms will become 1 and all the $Q_{1}Q_{3}, Q_{1}Q_{2}, Q_{2}Q_{3}$ will be zero.