For the first one we need all 3 vectors to represent u=(4 3 6), so it doenst matter which vector of the basis you remove.
For the second one we only need the first and the last vectors of the basis to represent u=(4 0 6), so removing the second basis-vector wont help. We have to remove the first or the last vector of the basis to geht 3 linear independent vectors.
Let the standard base is e1,e2,e3 (i)u=4*e1 + 3*e2 + 6*e3, and EVERY vector from the standard base can be removed, because of it’s linear combination of others. For example
e1= (1*u – 3*e2 – 6*e3)/4 = 0.25*u – 0.75*e2 – 1.5*e3
(ii) u = 4*e1 + 6*e3, and you can remove e1 or e3.
Geometric: every 2 vectors can be used to find only a point, belonging to the plain, described by them.
(i) u does not belong to a plane described by any two base vectors
(ii) u belongs to the plain described by e1 and e3
Hint: Here’s the geometric interpretation. Any two vectors in R^3 make a plane.
Add any vector which is not in that plane and you get a basis set (because the new vector is not a linear combination of the first two).