Find the real number $a$

Since $a$ is real, $\sqrt{a-11}$ must be real, so $a\geq11$ . Then $10-a\leq-1$ , so $\left| 10-a \right|=-(10-a)=a-10$ .

Then the equation becomes: $a-10+\sqrt{a-11}=a$

$\sqrt{a-11}=10$

$a=111$

I tried $a= 11$ and it didn't work, i'm lazy so I decided to try $a= x^2 + 11, \text{ for some } x \ge 1$ we get $\left| 10 - \left( x^2 + 11 \right) \right| + \sqrt{x^2 + 11 - 11} = x^2 + 11$ simplifying the above we get $x^2 + 1 + x = x^2 + 11 \Leftrightarrow x= 10$ Trying out our value for $x$ we see it holds and so $a= x^2 + 11= 111$

We see that $a\ge 11,$ so $10-a$ must be negative. Therefore, removing the absolute value signs, we have $-(10-a)+\sqrt{a-11}=a,$ which gets you to $\sqrt{a-11}=10 \implies \boxed{a=111}.$

Photomath bro 😏

Correct me please if I am wrong. Taking the absolute value into account the equation can be rewritten as
a-10 (abs) + (a-11)^1/2 = a
(a-11)^1/2 = a-a+10
squaring both sides
a-11 = 100
a = 111
To double check it putting this value back in equation
91 × 100^1/2 = 111
which is correct thus proving the solution to be correct.

|10-a|+ √(a-11 ) =a Can be rewritten as, √(a-11) = a -|10-a| Taking absolute value we get √a -11 = a + 10 -a => √ a-11 = 10 Squaring on both sides, We get a- 11 = 100 => a = 100 + 11 = 111