Correct option is (B) 2 only

Given equation is

\(\sqrt{2x-3}+\sqrt{3x-5}-\sqrt{5x-6}=0\)      _______________(1)

\(\Rightarrow\) \(\sqrt{2x-3}+\sqrt{3x-5}=\sqrt{5x-6}\)

\(\Rightarrow2x-3+3x-5\) \(+2\sqrt{(2x-3)(3x-5)}=5x-6\)     (By squaring both sides)

\(\Rightarrow\) \(2\sqrt{(2x-3)(3x-5)}\) \(=5x-6-5x+8\)

\(\Rightarrow\) \(2\sqrt{6x^2-19x+15}=2\)

\(\Rightarrow\) \(\sqrt{6x^2-19x+15}=1\)

\(\Rightarrow6x^2-19x+15=1\)      (By squaring both sides)

\(\Rightarrow6x^2-19x+14=0\)

\(\Rightarrow6x^2-12x-7x+14=0\)

\(\Rightarrow6x(x-2)-7(x-2)=0\)

\(\Rightarrow(x-2)(6x-7)=0\)

\(\Rightarrow x-2=0\;or\;6x-7=0\)

\(\Rightarrow x=2\;or\;x=\frac76\)

Let x = \(\frac{7}{6}\), 2x - 3 \(=2\times\frac{7}{6}-3\)

\(=\frac{7-9}3=\frac{-2}3<0\)

Hence, x = \(\frac{7}{6}\) is not in domain of given equation.

\(\therefore\) x = \(\frac{7}{6}\) is not a root of given equation.

Put x = 2 in equation (1)

\(\sqrt{2\times2-3}+\sqrt{3\times2-5}-\sqrt{5\times2-6}=0\)

\(\Rightarrow\) \(\sqrt{4-3}+\sqrt{6-5}-\sqrt{10-6}=0\)

\(\Rightarrow\) 1 + 1 - 2 = 0

\(\Rightarrow\) 0 = 0                   (Satisfied)

Hence, x = 2 is a root of given equation.

Correct option is B) 2 only