Let us solve
$$
\sqrt{x-6} + \sqrt{x-1} + \sqrt{x+6} = 9.
$$
Domain of solutions. It must be $x\ge 6$, $x\ge 1$ and $x\ge -6$, so $x\ge 6$. Also, $\sqrt{x-6}<\sqrt{x-1}<\sqrt{x+6}$ gives $3\sqrt{x-6}<9<3\sqrt{x+6}$ that implies:
$$
{\sqrt{x-6}<3<\sqrt{x+6}}\ \Rightarrow\ {x-6<9<x+6}\ \Rightarrow\ {3<x<15}.
$$
So it must be $6\le x<15$.
We have:
$$
\sqrt{x-6} + \sqrt{x+6} = 9 - \sqrt{x-1}.
$$
Squaring for the first time:
$$
x-6 + x+6 + 2\sqrt{x^2-36} = 81 + x-1 - 18\sqrt{x-1}
$$
$$
x + 2\sqrt{x^2-36} = 80 - 18\sqrt{x-1}
$$
$$
18\sqrt{x-1} + 2\sqrt{x^2-36} = 80 - x
$$
Squaring for the second time:
$$
324(x-1) + 4(x^2-36) + 72\sqrt{(x-1)(x^2-36)} = 6400 - 160x + x^2
$$
$$
324x - 324 + 4x^2 - 144 + 72\sqrt{x^3-x^2-36x+36} = 6400 - 160x + x^2
$$
$$
72\sqrt{x^3-x^2-36x+36} = 6868 - 484x - 3x^2
$$
Squaring for the third and last time:
$$
5184 (x^3 - x^2 - 36x + 36) = 47169424 + 234256x^2 + 9x^4 - 6648224x - 41208x^2 + 2904x^3
$$
$$
9 x^4 - 2280 x^3 + 198232 x^2 - 6461600 x + 46982800 = 0.
$$
With a bit of effort, we find:
$$
(x - 10) (9 x^3 - 2190 x^2 + 176332 x - 4698280) = 0
$$
where the first factor has one real root $x=10$ and the second factor has three distinct non-integer real roots, all of them greater than $69$. We conclude that the only acceptable answer is
$$
x=10.
$$
Easier method to search for an integer root, although not proving itself the unicity.
We have that $\sqrt{a}+\sqrt{b}$ or $\sqrt{a}+\sqrt{b}+\sqrt{c}$ are integers only if $a$, $b$ and $c$ are perfect squares.
Further, the sequence of differences of consecutive perfect squares $a_n=(n+1)^2-n^2$, $n\in\mathbb{N}$, is the sequence $b_n=2n+1$, $n\in\mathbb{N}$, of odd integers.
Three consecutive perfect squares, $\sqrt{x-6}$, $\sqrt{x-1}$, $\sqrt{x+6}$, whose difference are
$$
(x+6)-(x-1)=6+1=7
$$
and
$$
(x-1)-(x-6)=-1+6=5
$$
must be
$$
{x-6=4},\quad {x-1=9},\quad {x+6=16},
$$
all of them giving $x=10$.
