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I've got the coordinates of two points, the C in the pink square (see picture), and the coordinate of the OH. The coordinates are in 3D space; I also want to find the coordinate of CH3, but the only information I have are:

a. I know the length between C and CH3 b. I know the angle formed by CH3-C-OH (which is roughly 109 deg).

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I've tried to use the Rodrigues formula to rotate the vector formed by OH-C by 109 degrees to get the coordinate of CH3, but in this method, I'm wondering what's my $k$? Would $k$ be an arbitrary axis defined by the vector defined by the C in the pink square (see above picture) and the C above that C?

If it helps, these are the coordinates:

C (above pink box) 23.916 7.640 -15.538 C (within pink box) 24.525 7.246 -16.910 CH3 23.779 6.082 -17.562 OH 24.451 8.371 -17.791 ** this is what I'd like to predict, but how would I do this?

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    The locations of $C$ and $CH_3$ is not enough information to deduce the location of $OH$. Note that if we rotate the whole molecule about the $C-CH_3$ bond, we produce a whole bunch of possibilities of the position of the $OH$, all of which correspond to the same data. – 2017-02-13
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    Would knowing the coordinate of $C$ above the pink box help at all? – 2017-02-13
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    that might, actually. Can we assume that the C-C bond also has a 109 degree angle with the C-OH? – 2017-02-13
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    So basically we can sort of treat the problem as if it's a rotation about the C-C bond - that's why I was a bit confused! – 2017-02-13
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    I said a rotation about the C-CH3 bond. I don't think we can say that the CH3-C-C all lie on a straight line – 2017-02-13
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    C-CH3 and C-OH are all connected together, so we can assume that C->C(pink box)->CH3 are in one straight line, and C->C(pink box)->OH are also in a straight line – 2017-02-13
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    We can't, apparently You can verify numerically that the two Cs and the CH3 are not on colinear points. – 2017-02-13
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    Assuming that the two Cs and CH3 are colinear though, then how would we go about solving? Conceptually it sounds simple and there's a way to do it if you know the torsion angle about the C-C bound defined by N-C-C-OH but that's not always feasible – 2017-02-13
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    if the $C$s and $CH_3$s are colinear, then the second $C$ doesn't give us any information about where the OH might be relative to that C-CH3 axis. – 2017-02-13

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$k$ in Rodrigues formula is a normal to plane of rotation. Since you have only 2 points, you choose this plane among all planes which contain C-CH3 bond. So, just select arbitrary point in space (say, $A$), and then two vectors C-CH3 and C-A will lie in this plain. They cross-product will be orthogonal to the plain you have chosen, and thus you can rotate around it using this formula (after you normalize it by dividing by it's length, it's exactly vector $k$).