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?
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.