The point on the line $x+y=4$ that lie at a unit distance from the line $4 x+3 y=10$ is

Let the point on the line $x+y=4$ be $P(h, 4-h)$.

The distance $d$ of point $P(h, 4-h)$ from the line $4x+3y-10=0$ is given by the formula $d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.

Substituting the coordinates of $P$ and the line equation:

$1 = \dfrac{|4(h) + 3(4-h) - 10|}{\sqrt{4^2 + 3^2}}$

$1 = \dfrac{|4h + 12 - 3h - 10|}{\sqrt{16 + 9}}$

$1 = \dfrac{|h + 2|}{5}$

$|h + 2| = 5$

This gives two cases:

Case 1: $h + 2 = 5 \Rightarrow h = 3$. The point is $(3, 4-3) = (3, 1)$.

Case 2: $h + 2 = -5 \Rightarrow h = -7$. The point is $(-7, 4-(-7)) = (-7, 11)$.

Comparing with the given options, $(-7, 11)$ is present.

Answer: $(-7,11)$