#### Answer

$\approx 0.1732$

#### Work Step by Step

As we know that, $u=\dfrac{v}{|v|}$
Now, $u=\dfrac{\lt 1,1,1 \gt}{\sqrt{1^2+1^2+(1)^2}}=\lt \dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3} \gt$
Here, $dh=(\nabla h \cdot u) ds$
Thus, $dh=[\sqrt 3](0.1)\approx 0.1732$