<s>[INST] If ${\sqrt 3} - {\sqrt 2}, 4- {\sqrt 6}, p {\sqrt 3} - q {\sqrt 2}$ form a geometric progression, find the values of p and q. If ${\sqrt 3} - {\sqrt 2},  4- {\sqrt 6},  p{\sqrt 3} - q {\sqrt 2}$ form a geometric progression, find the values of p and q.
So I take the second term $4-{\sqrt 6} =( {\sqrt 3} - {\sqrt 2}) (r)$ , where r is the common ratio.
$4-{\sqrt 6} =( {\sqrt 3} - {\sqrt 2})( 2{\sqrt3} + {\sqrt2 })$
And found that the common ratio, r = $2{\sqrt3} + {\sqrt2 }$
To find the third term, I multiplied the second term with the common ratio.
$(4-{\sqrt 6})( 2{\sqrt3} + {\sqrt2 })=  p{\sqrt 3} - q {\sqrt 2}$ 
$8{\sqrt 3} + 4{\sqrt2} - 6 {\sqrt 2} - 2{\sqrt 6} =  p{\sqrt 3} - q {\sqrt 2}$ 
I am unable to proceed beyond this step. 
 [/INST] Note 
$$(4- {\sqrt 6})^2=({\sqrt 3} - {\sqrt 2})( p {\sqrt 3} - q {\sqrt 2})$$
or,
$$22-8\sqrt6 = 3p +2q -(p+q)\sqrt6$$
Therefore,
$$22= 3p +2q,\>\>\>\>\>p+q=8$$
Solve to obtain $p=6$ and $q=2$.
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