The distance between the line $x + 2 + t , y = 1 + t, z = -\frac{1}{2}-\frac{t}{2}$ and the plane $\vec{r}.(\hat{i}+ 2\hat{j}+6\hat{k})= 10 $, is

$\frac{9}{\sqrt{41}}$

The equation of the line is $\frac{x-2}{1}=\frac{y-1}{1}=\frac{z+\frac{1}{2}}{-1/2}=t.$

Clearly, it is possible to the plane $\vec{r}. (\hat{i} + 2\hat{j}+6\hat{k})= 10.$

So, required distance is equal to the length of perpendicular from $\left(2, 1, -\frac{1}{2}\right)$ to the plane $\vec{r}. (\hat{i} + 2\hat{j} + 6\hat{k})- 10 = 0 $ and is given by

$d = \begin{vmatrix} \frac{(2\hat{i}+\hat{j}-\frac{1}{2}\hat{k}).(\hat{i}+2\hat{j}+6\hat{k})-10}{\sqrt{1+4+36}} \end{vmatrix}=\frac{9}{\sqrt{41}}$