The equation of a straight line 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 ?
 A: 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 $s=-\dfrac43$.
A: The (unnormalized) vector from your fixed point to a candidate point on the line, ${\bf v}_1 (t)$ is:
$${\bf v}_1(t) = (3,-1,0) - \left( (2,1,1) + t(1,2,-1)\right)$$
The (unnormalized) vector of the line is just the difference of its end points:
$${\bf v}_2 = \left( (2,1,1) - (1,2,-1) \right) = (1,-1,0)$$
Set the dot product to zero, i.e., ${\bf v}_1(t) \cdot {\bf v}_2$:
$$(1,-1,0) \cdot (1,-2,-1) + (1, -1,0) \cdot t (1,2,-1) = 0$$
Solve to find $t=3$.  Then plug in to find the point.
A: 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.
