3. Solve the equation $\sqrt{6-x}+\sqrt{x-4}=x^{2}-10 x+27$.

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3. Given $4 \leq x \leq 6$. Find the maximum of the function $f(x)=\sqrt{6-x}+\sqrt{x-4}$

$f^{\prime}(x)=-\frac{1}{\sqrt{6-x}}+\frac{1}{\sqrt{x-4}} \Rightarrow f^{\prime}(x)=0 ; \sqrt{4-x}=\sqrt{6-x} \Rightarrow x=5$. When $x=5$, $f(x)$ has a maximum value of 2. On the other hand, $x^{2}-10 x+27=(x-5)^{2}+2 \Rightarrow \min \varphi(x)=x^{2}-10 x+27=2$ when $x=5$. Therefore, equality is possible only when $x=5$. Answer: $\{5\}$.