Given :

\(\frac{3}{{\text{x}}-2}<2\), x ∈ R

Subtracting 2 from both the sides in the above equation,

\(\frac{3}{{\text{x}}-2}-2\)<2-2

\(\frac{3-2({\text{x}}-2)}{{\text{x}}-2}<0\)

\(\frac{3-2{\text{x}}+4}{{\text{x}}-2}\) <0

\(\frac{7-2{\text{x}}}{{\text{x}}-2} <0\)

Signs of 7 – 2x:

7 - 2x = 0 → x = \(\frac{7}{2}\)

(Subtracting by 7 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)

 7 - 2x < 0 → x > \(\frac{7}{2}\)

(Subtracting by 7 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)

7 – 2x > 0 → x < \(\frac{7}{2}\)

(Subtracting by 7 on both the sides, then multiplying by -1 on both the sides and then dividing both the sides by 2)

Signs of x – 2: 

x – 2 = 0 → x = 2 (Adding 2 on both the sides) 

x – 2 < 0 → x < 2 (Adding 2 on both the sides) 

x – 2 > 0 → x > 2 (Adding 2 on both the sides)

Zeroes of denominator:

x – 2 = 0 → x = 2

At x = 2, \(\frac{7-2{\text{x}}}{{\text{x}}-2} \) is not defined

intervals satisfying the condition: <0

x<2 and x > \(\frac{7}{2}\)

Therefore,

x ∈ (-∞,2) U \(\big(\frac{7}{2}, ∞\big)\)