Is there non real number of x that sufficient for this $11-\sqrt 7 x = 4x - 10$ 
Is there non real number of x that sufficient for $11-\sqrt 7 x = 4x - 10$ and $|\sqrt 7 x - 11 | = 4x - 10$

There is a solution which is real number.

$|anything|$ is always a positive real number.
So $|anything| = 4x - 10$ will only have real solutions (if any solutions at all).
Also for complex numbers, $x$ we tend not to use the notation  $\sqrt{x}$ as it is ambiguous.
I'm going to assume you meant $\sqrt{7x}$ (which is the square root of $7x$) and not $\sqrt{7}x$ (which is $x$ times $\sqrt 7$)
$11 - \sqrt{7x} = 4x -10$
$4x + \sqrt 7\sqrt x -21 = 0$
$\sqrt{x} =\frac {-\sqrt 7 \pm \sqrt{7+4*21*4}}8=$
$\frac {-\sqrt 7\pm \sqrt {342}}8$
Note:  $\frac {-\sqrt 7- \sqrt {342}}8$ is negative while $\frac {-\sqrt 7+ \sqrt {342}}8$ is positive.  But both are real.
Even if we use $\sqrt{x}$ to allow for other than non-negative real numbers and to apply for any $k$ so that $k^2 = x$ (which is not what we do... at least not for real number-- and if we use complex numbers we avoid the notation $\sqrt{}$ altogether:  but for the sake of being thourough, I will pretend we can allow $\sqrt x < 0$.... which we really can not do....) we will nave
$x = (\frac {-\sqrt 7\pm \sqrt {342}}8)^2$ which can only be a positive real number.
Now if we have $|11-\sqrt{7x}|\ne 11- \sqrt{7x}$ and as $11-\sqrt{7x}$ is real mus mean $|11-\sqrt{7x}| = \sqrt{7x} - 11 = 4x-10 = 11-\sqrt{7x}$ which is impossible.
So $x = (\frac {-\sqrt 7\pm \sqrt {342}}8)^2$.
BUT as we DON'T define $\sqrt{x}$ as any $k$ so that $k^2 =x$ but only as the non-negative real $k$ so that $k^2 = x$ we have
$x= (\frac {-\sqrt 7+ \sqrt {342}}8)^2$ is the only real solution.
