Unknown huge Volatge drop over a resistor
One of the most basic electrical engineering formulas is Ohm's Law $$E = iR $$

With current (i) in amps and (R) in ohms, for your circuit the voltage drop across the resistor is $$E = iR = .068 \times 150 = 10.2 \text{ volts} $$ And, of course, I'll just bet that your resistor is not a 1% unit, but perhaps a 10% tolerance. Then the actual resistance could be 165 ohms, for a drop of 11.2 volts. Add 0.7 volts for the diode, and 4 volts sounds just about right.

Furthermore, you need to read the specs on your multimeter. It will measure current by putting a resistor in series and measuring the voltage across the resistor. There is no reason to assume that the shunt resistor your meter uses is negligeable. Let's assume that it is 10 ohms. Then the effective total resistance is 175 ohms for a voltage drop of 11.9 volts with the meter in place, for an output voltage of 4.1 volts, and less if you count the diode.

So you need to find the manual for your meter (or look it up on the web) and determine just what your current measurement burden (that's the technical term) is. You also need to use the meter to measure your 150 ohm resistor. Additionally, you need to measure your 16 volt supply to make sure it is not actually putting out something like, say, 15 volts. And finally, you need to measure the current without the resistor in place. This would not be important if it were not for your statement "it dropped down to 4 voltage so the LP 2950 cant regulate the voltage properly." According to the spec sheet, at the maximum rated 100 mA load, the dropout voltage should be less than 0.5 volts, so a 3.3 volt unit should be quite happy with a 4 volt input. So, as user3528438  commented, it seems very likely that you are drawing too much current for the regulator. If so, the obvious next step is to provide a separate regulator for each unit.