How can we find the line? I want to find the line that passes through $(3,1,-2)$ and intersects at a right angle the line $x=-1+t, y=-2+t, z=-1+t$. 
The line that passes through $(3,1,-2)$ is of the form $l(t)=(3,1,-2)+ \lambda u, \lambda \in \mathbb{R}$ where $u$ is a parallel vector to the line. 
There will be a $\lambda \in \mathbb{R}$ such that $3+ \lambda u_1=-1+t, 1+ \lambda u_2=-2+t, -2+ \lambda u_3=-1+t$. 
Is it right so far? How can we continue?
 A: Hint :
Equation of line passing through $(3,1,-2)$ is given by $$\frac{x-3}{l}=\frac{y-1}{m}=\frac{z+2}{n}.$$This line is perpendicular to the line $\displaystyle \frac{x+1}{1}=\frac{y+2}{1}=\frac{z+1}{1}$.
A: The line you're given is 
$$(-1,-2,1)+t(1,1,1)$$
Thus, as you said, you want a line $\;(3,1,-2)+\lambda u\;$ , with $\;u\perp(1,1,1)\;$ , so if $\;u=(a,b,c)\;$ , you need
$$u\cdot(1,1,1)=a+b+c=0$$
From here it is easier, I beleive.
A: A normal vector to the plane passing through $A(3,1,-2)$ perpendicular to the given line is the directing vector $(1,1,1)$ of this line. If it passes through the point $(3,1,-2)$, a cartesian equation of the plane is thus:
$$x+y+z=3+1-2=2.$$
It intersects the line at the point  $B(t-1,t-2,t-1)$ which satisfies this equation, i.e. $3t-4=2$, whence $t=2$. Thus the sought for line has directing vector $\overrightarrow{BA} (2,1,-3) $ and a parametric representation of this line is, vectorially, $M=A+u\,\overrightarrow{BA} $, i.e.
$$x=3+2u,\enspace y=1+u,\enspace z=-2-3u.$$
