Find, correct to the nearest degree, the three angles of the triangle with the given...

A(1,0,-1), B(3,-2,0), C(1,3,3)
Find vectors to represent the sides.
$\overrightarrow{AB}=<3-1,-2-0,0-(-1)>=<2,-2,1>$
$\overrightarrow{AC}=<1-1,3-0,3-(-1)>=<0,3,4>$
$\overrightarrow{BC}=<1-3,3-(-2),3-0>=<-2,5,3>$
Find the magnitudes of the vectors.
$|\overrightarrow{AB}|=\sqrt{2^2+(-2{)}^2+1^2}=3$
$|\overrightarrow{AC}|=\sqrt{0^2+3^2+4^2}=5$
$|\overrightarrow{BC}|=\sqrt{(-2{)}^2+5^2+3^2}=\sqrt{38}$
Use the dot product to find the angle
$\cos \theta =\frac{a\cdot b}{|a||b|}\to \theta =\arccos \frac{a\cdot b}{|a||b|}$
$\overrightarrow{AB},\overrightarrow{AC}:\theta =\arccos \frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|}$
$=\arccos \frac{2(0)+(-2)(2)+1(4)}{3(5)}=\arccos \frac{-2}{15}\approx {97.67}^{\circ }$
$\overrightarrow{AB},\overrightarrow{BC}:\theta =\arccos \frac{\overrightarrow{AB}\cdot \overrightarrow{BC}}{|\overrightarrow{AB}||\overrightarrow{BC}|}$
$=\arccos \frac{2(-2)+(-2)(5)+1(3)}{3(\sqrt{38})}=\arccos \frac{-11}{3\sqrt{38}}\approx ={126.5}^{\circ }$
Because of the direction $\overrightarrow{BC}$ is pointing relative to $\overrightarrow{AB}$ this is the angle outside the triangle, and we should find the supplementary angle.
${180}^{\circ }-{126.5}^{\circ }={53.5}^{\circ }$
If we used $-(\overrightarrow{BC})=\overrightarrow{CB}$ in the formula (just negative the numerator) we would have gotten the correct angle on the first try.
The 3rd angle should be what's left over after subtracting from ${180}^{\circ }$
${180}^{\circ }-{97.67}^{\circ }-{53.5}^{\circ }={28.83}^{\circ }$
You can confirm using the formula again:
$\overrightarrow{AC},\overrightarrow{BC}:\theta =\arccos \frac{\overrightarrow{AC}\cdot \overrightarrow{BC}}{|\overrightarrow{AC}||\overrightarrow{BC}|}$
$=\arccos \frac{0(-2)+3(5)+4(3)}{5(\sqrt{38})}=\arccos \frac{27}{5\sqrt{38}}\approx {28.84}^{\circ }$
We'll round everything to I decimal place so they add up to 180.
Result:
Angle at vertex $A:{97.7}^{\circ }B:{53.5}^{\circ }C:{28.8}^{\circ }$