I need to find the equation of a plane that passes through the point with position vector $A = 5i + j + 3k$ and the line $l_1 = 2i-8j-17k + \lambda(i-3j-4k)$.
I found the normal by finding the cross product of $A * [(5i+j+3k) - (2i-8j-17k)]$ which turned out to be $n_1=-24i-32j+18k$. Then I used $r \cdot n = a\cdot n \to(5i+j+3k) \cdot (-24i-32j+18k)=-98$.
However, it turns out that the above value should be $49$ rather than $-98$. I realize that it can be divided by -$2$ to get the answer; however, I don't understand why this happens.
$-98$ is correct if you use $n_1=-24i-32j+18k$ as normal vector. Then the equation of the plane is $$-24x-32y+18z=-98\tag{1}.$$ But you can divide the normal vector by $-2$, obtaining $n_2=12i+16j-9k$. For such normal vector the correct value is $49$ and the equation of the plane is $$12x+16y-9z=49\tag{2}.$$ Of course the equations (1) and (2) are equivalent and they represents the same plane.