Finding distance from point to line which is perpendicular to another line Find the distance of the point $(1,1,1)$ from $x+y+z=1$ measured perpendicular to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$

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 $Q\left(-1,-\frac{1}{3},\frac{7}{3}\right)$, and the required distance is:
$$PQ=\sqrt{{(2)}^{2}+{\left(\frac{4}{3}\right)}^{2}+{\left(-\frac{4}{3}\right)}^{2}}=\sqrt{\frac{68}{9}}$$
$$\Rightarrow PQ=\frac{2\sqrt{17}}{3}$$ 
