Recall that for two lines to be skew they must satisfy two conditions:
1) They must not be parallel.
2) They must not intersect.
We shall check each condition individually.
The general vector equation of a line is given by
r = a + kb,
where a is the position vector of some point on the line, k is a scalar, and b is the direction vector of the line. The direction vector of the line, as the name suggests, dictates in what direction the line travels; b tells us how the line is orientated in space.
For two lines to be parallel, the direction vector of one line must be equal to some scalar multiple of the second line. However, for our two lines, it is clear that there exists no scalar k for which
(-1,2,2) = k(1,3,5).
Thus, the two lines cannot be parallel.
Let us assume that the two lines do in fact intersect. In other words, that
(1,4,1)+s(-1,2,2) = (2,8,2)+t(1,3,5)
for some numbers s and t.
This vector equation leads to three simultaneous equations:
1-s = 2+t (1), 4+2s = 8+3t (2), 1+2s = 2+5t (3).
If we add 2 times Eq. (1) to Eq. (2), we get that
t = -6/5.
If we substitute this value of t into, say, Eq. (3), we get that
s = -5/2.
However, subsituting both of these values into Eq. (2) yields a contradiction. The LHS gives
4+2(-5/2) = 4-5 = -1,
whereas the RHS gives
8+3(-6/5) = 22/5.
Clearly, then, the LHS is not equal to the RHS; the system of equations is inconsistent, and so the lines do not intersect.
We have shown that the given lines satisfy both of the necessary conditions to be classified as skew. The lines are therefore skew, as required.