The sum of squares of the abscissas of all the points on the line $x+y=4$ that lie at a unit distance from the line $4x+3y-10=0$ is

Any point on the line $x+y=4$ can be taken as $\left(x_1,4-x_1\right).$

As it is at a unit distance from the line $4x+3y-10=0$, we get,

 $\frac{\left|4x_1+3\left(4-x_1\right)-10\right|}{\sqrt{\left(4^2+3^2\right)}}=1$
$\Rightarrow \left|x_1+2\right|=5\Rightarrow x_1+2=\pm 5$

$\Rightarrow x_1=3$ or $-7$
$\therefore$ Required sum of the squares $=9+49=58$