Question 865766
NOTE: Check my calculations, because it is way past my bedtime.
 
The diagonals are
{{{drawing(20,20,-0.2,1.2,-0.2,1.2,arrow(-0.1,1,1.1,1),locate(0,0.9,OB))}}}{{{"="}}}{{{drawing(20,20,-0.2,1.2,-0.2,1.2,arrow(-0.1,1,1.1,1),locate(0,0.9,OA))}}}{{{"+"}}}{{{drawing(20,20,-0.2,1.2,-0.2,1.2,arrow(-0.1,1,1.1,1),locate(0,0.9,OC))}}}{{{"="}}}{{{(i+3j+3k)+(3i-j+k)=4i+2j+4k}}}
AND
{{{drawing(20,20,-0.2,1.2,-0.2,1.2,arrow(-0.1,1,1.1,1),locate(0,0.9,AC))}}}{{{"="}}}{{{drawing(20,20,-0.2,1.2,-0.2,1.2,arrow(-0.1,1,1.1,1),locate(0,0.9,OC))}}}{{{"-"}}}{{{drawing(20,20,-0.2,1.2,-0.2,1.2,arrow(-0.1,1,1.1,1),locate(0,0.9,OA))}}}{{{"="}}}{{{(3i-j+k)-(i+3j+3k)=2i-4j-2k)}}}
So {{{OB=sqrt(4^2+2^2+4^2)=sqrt(16+4+16)=sqrt(36)=6}}} and
{{{AC=sqrt(2^2+4^2+2^2)=sqrt(4+16+4)=sqrt(24)=2sqrt(6)}}}
The diagonals bisect each other and split parallelogram OABC into 4 triangles.
Two sides of each triangle will be the halves of the two diagonals.
That is, two sides will have lengths of {{{6/2=3}}} and {{{2sqrt(6)/2=sqrt(6)}}} .
The remaining side of each triangle will be one side of the parallelogram, and not all sides of the parallelogram are the same length. (It is not a rhombus).
The parallelogram's side lengths are:
{{{OA=sqrt(1^2+3^3+3^2)=sqrt(1+9+9)=sqrt(19)}}}
{{{OC=sqrt(3^3+1^2+1^2)=sqrt(9+1+1)=sqrt(11)}}}
The angle between the sides measuring {{{3}}} and {{{sqrt(6)}}} will be smaller in the two triangles where it is opposite the shorter parallelogram side measuring {{{sqrt(11)}}} .
We need the measure of that angle, that I will call {{{theta}}} .
The law of cosines, applied to that triangle says that
{{{(sqrt(11))^2=(3)^2+(sqrt(6))^2)-2(3)(sqrt(6))cos(theta)}}}
{{{11=9+6-6sqrt(6)cos(theta)}}}
{{{6sqrt(6)cos(theta)=9+6-11}}}
{{{6sqrt(6)cos(theta)=4}}}
{{{cos(theta)=4/6sqrt(6)}}}
{{{cos(theta)=2/3sqrt(6)}}}
{{{cos(theta)=2sqrt(6)/(3*6)}}}
{{{cos(theta)=sqrt(6)/9}}} ---> {{{theta=74.2^o}}}