1) As we need to find a plane perpendicular to $u_1$, we should start by calculating $u_1$:
This is the vector which is shortest, so it will be shortest in the 2-norm or Euclidean norm:
(4 + t)^2 + (3+2t)^2 + 2^2 = 16 + 8t + t^2 + 9 + 12t + 4t^2 + 4 \\
= 5t^2 + 20t + 29\\
Now these quadratics have a global minimum, I'll let you do that since this looks very homework-question-y.
After that, say $u_1 = (v_1,v_2,v_3)$, we want to find a plane which is perpendicular. I refer you to this question. The answer provided there does a good job of explaining why this is the case.
2) We know a point in plane 1 (the origin!), so we this is the same as asking what the shortest distance the second plane has to the origin point.
Now, in our heads, we should have a triangle $OXP$, with one point at $0$, another at $X$ the shortest point in plane 2, and a general point $P$ in plane 2.
From here, $\angle OXP$ is $90^\circ$, as it is the shortest point. The question is to find $||OX||$.
Again as this is a homework style question, I'll let you finish it off.
EDIT: I realise that this might not be the best solution to it, but once you understand it, this is the best solution - if that makes sense.
Think about the problem geometrically, now if a planes (2-d things) are parallel in a 3-d space, then they both have a 'normal' vector, which is normal to the two vectors spanning the plane.
E.g. if $\Pi = \bar w + \bar v_1 t + \bar v_2 r$, then there is a normal vector $\bar n$ such that $\bar n.\bar v_1 = \bar n.\bar v_2 = 0$. (Perpendicular to the other two)
If two planes are parallel, then the normal vector is off by a scalar.
Now, again, the plane is always normal to this point $\bar n$, which means, for every $\bar x$ in the plane, we have that $\bar x. \bar n = \bar w . \bar n$ as the $\bar v_1, \bar v_2$ are both perpendicular to $\bar n$.
Now what is $\bar w$, well it's the vector from $0$ to $w$ as coordinates. In particular, it has no special properties other than living in the plane.
So we have $$\bar n . \bar x = n_1x_1 + n_2x_2 + n_3x_3 = d_1.$$
Realising what this means, we have that $a/A = b/B = c/C$. And we also have that the $d_1$ has some relation with the shortest distance to the origin (think about the triangle i was making above again). I'll add another hint if this is still not a complete result, but you should have it from here.
If not try youtube, this is literally the first that came up when searching, haven't seen it, but imagine it has what you are after.
P.S. If you're looking to be ready for university maths, a good idea might be to use the siklos booklets, or here. Or for a slightly more abstract start, try abstract algebra in a wonderful book that is very gentle to introduce concepts.