Is this line uniquely determined? Let $\alpha: x-y+2z-2=0$ and $\beta: x-2y-2z+3=0$ be two planes in $\mathbb{R}^3$. I am asked to find a line $d_1\subset \alpha$ such that $d_1$ is perpendicular on the line $\alpha \cap \beta$ and that passes through the point $(1, 1, 1)$.
Here's what I did: I intersected the planes $\alpha$ and $\beta$ and I got that $\alpha\cap \beta: \frac{x-7}{-6}=\frac{y-5}{-4}=\frac{z}{1}$.
Let $u=(a, b, c)$ be a direction vector of $d_1$. Since $d_1 \perp \alpha \cap \beta$, we must have  $-6a-4b+c=0$. Because $d_1 \subset \alpha$, it follows that $a-b+2c=0$ ($u$ must be in $\alpha$'s direction). If I solve these two equations together, I get that $b=-\frac{13}{7}a$ and $c=-\frac{10}{7}a$. As a result, I can take the direction vector to be $(7, -13, -10)$.
I got that $d_1:\frac{x-1}{7}=\frac{y-1}{-13}=\frac{z-1}{-10}$. This is indeed a line contained in $\alpha$. I wonder, however, if the conditions I imposed guarantee that the result is going to be correct. My question comes from that fact that the condition that $u$ is in $\alpha$'s direction only implies that my vector is parallel to the plane. Is that perpendicularity enough to make sure that what I will get out of the system of equations is what I want?
 A: You have already pointed out that the equation $a-b+2c=0$ guarantees that $d_1$ is parallel to the plane $\alpha$. Note that $d_1$ also passes through $(1,1,1)$, which is a point on $\alpha$. This guarantees that the line $d_1$ lies on the plane $\alpha$, and not just parallel to it.
If that condition had not been given, we would have got a family of parallel lines with fixed direction cosines, which you obtained.
A: Please note that the point $(1, 1, 1)$ is in the plane $\alpha$. So if the line is parallel to the plane and the point is on the line, the line must be in the plane. Just an alternative approach to find the equation of the line instead of building system of equations,
Normal vector to the plane $\alpha$ is $(1, -1, 2)$ and the direction vector of intersecting line ($\alpha \cap \beta$) is $(-6, -4, 1)$. So the direction vector of the line $d_1$ in the plane which is perp to $\alpha \cap \beta$, will be given by cross product of $(1, -1, -2)$ and $(-6, -4, 1)$. That gives us direction $(7, -13, -10)$ and equation of the line $d_1$ is,
$\dfrac{x-1}{7} = \dfrac{y-1}{-13} = \dfrac{z-1}{-10}$
