Pure formula based

The equation of the straight line perpendicular to the given line $\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$ and passing through the point $P(1,1,1)$ is:

$\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}$

where, $2a+3b+6c=0$ . Clearly, $(a,b,c)=(3,2,-2)$ is a solution.

Let $\frac{x-1}{3}=\frac{y-1}{2}=\frac{z-1}{-2}=t$

Then, any point on this line can be written as $(1+3t,1+2t,1-2t)$ . Let this line intersect the plane $x+y+z=1$ at $Q(1+3t,1+2t,1-2t)$ . Then,

$(1+3t)+(1+2t)+(1-2t)=1 \Rightarrow t=-\frac{2}{3}$

So, the point of intersection is $Q(-1,-\frac{1}{3},\frac{7}{3})$ , and the required distance is:

$PQ=\sqrt{{(2)}^{2}+{(\frac{4}{3})}^{2}+{(-\frac{4}{3})}^{2}}=\sqrt{\frac{68}{9}}$

$\Rightarrow PQ=\frac{2\sqrt{17}}{3} \approx \boxed{2.749}$