Let the domain of the function $$f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2)$$ be $$(a, b)$$. If $$\int_0^{b-a} [x^2]\,dx = p - \sqrt{q} - \sqrt{r}$$, $$p, q, r \in \mathbb{N}$$, $$\gcd(p, q, r) = 1$$, then $$p + q + r$$ is equal to

The given function is $$f(x)=\log_{2}\!\Bigl[\log_{4}\!\bigl[\log_{6}(3+4x-x^{2})\bigr]\Bigr]$$.
For the domain, the argument of each logarithm must be positive.

Step 1 (innermost log)
$$3+4x-x^{2}\; \gt \;0$$ is needed for $$\log_{6}$$ to exist.

Step 2 (middle log)
For $$\log_{4}[\;]$$ we again need its argument positive:
$$\log_{6}(3+4x-x^{2})\; \gt \;0 \;\;\Longrightarrow\;\; 3+4x-x^{2}\; \gt \;6^{0}=1 .$$

Step 3 (outermost log)
Finally, $$\log_{2}[\;]$$ demands
$$\log_{4}\!\bigl[\log_{6}(3+4x-x^{2})\bigr]\; \gt \;0 \;\;\Longrightarrow\;\; \log_{6}(3+4x-x^{2}) \; \gt \;4^{0}=1 \;\;\Longrightarrow\;\; 3+4x-x^{2} \; \gt \;6 .$$

Simplify the last inequality:
$$3+4x-x^{2}\; \gt \;6 \;\;\Longrightarrow\;\; -x^{2}+4x-3 \; \gt \;0 \;\;\Longrightarrow\;\; x^{2}-4x+3 \; \lt \;0 \;\;\Longrightarrow\;\; (x-1)(x-3) \; \lt \;0.$$ The solution of $$(x-1)(x-3)\lt0$$ is $$1\lt x\lt3.$$
Hence the domain is $$(a,b)=(1,3),$$ so $$b-a=2.$$

We now evaluate the definite integral
$$\int_{0}^{\,b-a}[x^{2}]\,dx=\int_{0}^{2}[x^{2}]\,dx,$$ where $$[x^{2}]$$ denotes the greatest-integer (floor) function.

Break the interval $$[0,2]$$ wherever $$x^{2}$$ crosses an integer:

• $$0\le x\lt1:\;x^{2}\in[0,1)\;\Rightarrow\;[x^{2}]=0$$
• $$1\le x\lt\sqrt2:\;x^{2}\in[1,2)\;\Rightarrow\;[x^{2}]=1$$
• $$\sqrt2\le x\lt\sqrt3:\;x^{2}\in[2,3)\;\Rightarrow\;[x^{2}]=2$$
• $$\sqrt3\le x\le 2:\;x^{2}\in[3,4]\;\Rightarrow\;[x^{2}]=3$$ for $$x\lt2,$$ (the single point $$x=2$$ where $$[4]=4$$ has zero measure and does not affect the integral).

Compute the area on each sub-interval:

$$\begin{aligned} \int_{0}^{1}0\,dx &= 0,\\ \int_{1}^{\sqrt2}1\,dx &= (\sqrt2-1),\\ \int_{\sqrt2}^{\sqrt3}2\,dx &= 2(\sqrt3-\sqrt2),\\ \int_{\sqrt3}^{2}3\,dx &= 3(2-\sqrt3). \end{aligned}$$

Add them: $$\begin{aligned} 0 &+ (\sqrt2-1) + 2(\sqrt3-\sqrt2) + 3(2-\sqrt3) \\ &= \sqrt2-1 + 2\sqrt3-2\sqrt2 + 6-3\sqrt3 \\ &= 5 - \sqrt2 - \sqrt3. \end{aligned}$$

Thus $$\int_{0}^{2}[x^{2}]\,dx = 5 - \sqrt2 - \sqrt3.$$
Comparing with $$p-\sqrt{q}-\sqrt{r},$$ we have $$p=5,\;q=2,\;r=3.$$ Because $$\gcd(5,2,3)=1,$$ the required sum is $$p+q+r = 5+2+3 = 10.$$

Answer (Option A): 10