This is a question which puzzled our entire math class including our teacher, I'm referring to part (b), we're fine with part (a). We don't understand the reason for taking the dot product and the significance of what it produces. Any explanation of this question would be much appreciated.
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Hint:
A plane equation can be written as $\vec n\cdot \vec p=d$, where $n$ is a unit vector perpendicular to the plane and $d$ the shortest distance to the origin.
Given a general equation $ax+by+cz+d=0$, putting it in the above form isn't an issue (don't confuse the $d$'s).
As the planes are parallel, they share the normal and you need to solve
$$|d_2-d_1|=|\vec n\cdot\vec p_2(\alpha)-\vec n\cdot\vec p_1|$$ for $\alpha$.
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First, it's important to notice that a point M (x, y, z) has an unique associated vector with the origin equals the origin of coordinate system and the extremity in M.
The condition for a point M to belong to a plain $ax + by + cz = d$ is that the scalar product between the vector the point M is associated with and the plane's normal vector is equal d. This follows directly from scalar product definition and the fact that the plane's normal vector is (a,b,c).
They used the scalar product condition for $(\alpha,1,1)$ belongs to $\pi_2$ plane.
Evaluate with the formula of distance between plane $\;\pi_1\;$ and point $\;(\alpha,1,1)\in\pi_2\;$ . Since the planes are parallel you can take any point in one of them and calculate its distance to the other one:
$$\frac{|3\alpha-10-4+13|}{\sqrt{3^2+10^2+4^2}}=\frac{|3\alpha-1|}{5\sqrt5}\stackrel{\text{given}}=\frac1{\sqrt5}\iff$$
$$\;\implies\;\;|3\alpha-1|=5$$
and now solve the easy pair of equations to find out the possible values of $\;\alpha\;$ .
Slightly worrying also your teacher was puzzled by this, yet we all have some awful moments here and there.