Question: Find the equation of the straight line passing through the point $(3 , -1 ,0)$ and intersects the straight line $r=(2,1,1) + t(1,2,-1)$ orthogonally.

Since the two straight lines are intersecting orthogonally, their direction vectors are perpendicular

$$d\_1 = (1,2,-1) , d\_2 = (a ,b ,c) $$

$$d\_1\cdot d\_2 = 0 = a +2b -c $$

Let $a=b = 1$, then $c = 3 , d\_2 = (1,1,3)$

Then the required equation is $r = (3,-1 ,0) + t(1,1,3)$
Is my solution correct ?

Answer: 
There are many lines containing the point $(3,-1,0)$ with direction vector perpendicular to $(1,2,-1)$ 

-- they are in a plane -- but you need one that intersects the given line, 

and the one you found does not. 

As you correctly indicated, the direction perpendicular to $(1,2,-1)$ is $(a,b,c)$, with $a+2b-c=0$. 

It is fine to scale $(a\,',b\,',c\,')$ to $(1,b,c)$, but you were incorrect to set $b=1$ too.

So we need the line $(2,1,1)+t(1,2,-1)$ to intersect $(3,-1,0)+s(1,b,c)$. 

Equating components yields equations $2+t=3+s, 1+2t=-1+sb,$ and $ 1-t=sc$. 

We also have $1+2b-c=0$. 

Solving these four equations in four unknowns (details available upon request) yields $t=-\dfrac13$, $s=-\dfrac43$, and $b=c=-1$. Thus, the desired line is $(3,-1,0)+s(1,-1,-1);$ 

it intersects the other line when $s=-\dfrac43$.
HINT: If the lines are to meet orthogonally, they meet at the point on $r\_1(t) = (2,1,1) + t(1,2,-1)$ closest to $(3,-1,0)$. So your first task is to minimise the distance between the point and the line. That is, minimise
$$\sqrt{(-1+t)^2+(2+2t)^2+(1-t)^2}.$$
(You might as well ignore the square root, and if you know calculus it should help you find the minimum.)
Once you have the value of $t$, plug it into $r\_1(t)$ to find the desired intersection point.
Once you have that you can set any $s \neq 0$ you like in $r\_2(s) = (3,-1,0) + s(a,b,c)$ and solve for $a,b,c$, scale it if you like. Check that $a + 2b - c = 0$ like you suggested. If so, and the lines do indeed intersect, you have it.