Can someone explain to me how to solve this problem and please show all work

VV lllUUJ. AULI 1. For each real number 3, let 133 be the point (1 s, 1 + s, 2 28). For each real number If, let qt be the point (4,1 + t, 2t). This problem concerns the lines 61 = {p3 : s E R} and 2 = {qt : t E R}. (a) Let f(s,t) = dist(ps,qt)2. Calculate Vf(s,t). (b) Find the point (so, to) where this gradient vanishes. (c) Show that the line segment joining the points 1330 and qto is perpendicular to 1 and 62 (d) Explain why the distance from p80 to qto is the minimum distance from 1 to 62, in terms of the segment you found in part c. (Hint: If W is any vector joining a point on 61 to a point on 2, consider its projection onto the segment 7 joining p30 to qto.)