Equivalent conductance of $BaC{{l}_{2}},\text{ }{{H}_{2}}S{{O}_{4}}$and $HCl$ are ${{x}_{1}},\,{{x}_{2}}\text{ }and\,{{x}_{3}}$ $Sc{{m}^{2}}equi{{v}^{-1}}$ at infinite dilution. If specific conductance of saturated $BaS{{O}_{4}}$ solution is of y $Sc{{m}^{1}}$ then ${{K}_{sp}}$ of $BaS{{O}_{4}}$ is:
(a)- $\dfrac{{{10}^{3}}y}{2({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}$
(b)- $\dfrac{{{10}^{6}}{{y}^{2}}}{4{{({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}^{2}}}$
(c)- $\dfrac{{{10}^{6}}{{y}^{2}}}{2{{({{x}_{1}}+{{x}_{2}}-{{x}_{3}})}^{2}}}$
(d)- $\dfrac{{{x}_{1}}+{{x}_{2}}-2{{x}_{3}}}{{{10}^{6}}{{y}^{2}}}$

Hint: We can calculate the equivalent conductance of barium sulfate by adding the equivalent conductance of barium chloride and sulfuric acid and then subtracting it with the twice of equivalent conductance of hydrochloric acid. The solubility product of barium sulfate can be calculated by the square of the concentration of ions.

Complete step by step answer:
First, we have to calculate the equivalent conductance of barium sulfate ($BaS{{O}_{4}}$), and we are given the equivalent conductance of $BaC{{l}_{2}},\text{ }{{H}_{2}}S{{O}_{4}}$and $HCl$ are ${{x}_{1}},\,{{x}_{2}}\text{ }and\,{{x}_{3}}$ $Sc{{m}^{2}}equi{{v}^{-1}}$. So, with these we can calculate the equivalent conductance of $BaS{{O}_{4}}$as: adding the equivalent conductance of barium chloride and sulfuric acid and subtracting the twice of equivalent conductance of hydrochloric acid.
${{\Delta }_{eq}}(BaS{{O}_{4}})={{\Delta }_{eq}}(BaC{{l}_{2}})+{{\Delta }_{eq}}({{H}_{2}}S{{O}_{4}})-2{{\Delta }_{eq}}(HCl)$
By putting the values, we get
${{\Delta }_{eq}}(BaS{{O}_{4}})=({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})$ $Sc{{m}^{2}}equi{{v}^{-1}}$
We also know that the equivalent conductivity is equal to the product of specific conductance and one gram equivalent of the electrolyte, the equation is given:
${{\Delta }_{eq}}=\kappa \text{ x }V$
We know the volume of the electrolytic solution is taken in $c{{m}^{3}}$ so,
$V=\dfrac{1000}{N}$
Therefore,
${{\Delta }_{eq}}=\kappa \text{ x }\dfrac{1000}{N}$
Where N is the normality,
So, the normality can be calculated from this equation as,
$N=\kappa \text{ x }\dfrac{1000}{{{\Delta }_{eq}}}$
Given the specific conductance of barium sulfate is y, and putting all the values in the equation, we get:
$N=\text{ }\dfrac{y\text{ x }{{10}^{3}}}{({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}$
Now, this normality can be converted into molarity (concentration) by dividing it with 2 because the valency of $BaS{{O}_{4}}$ is 2.
$M=\text{ }\dfrac{y\text{ x }{{10}^{3}}}{2({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}$
The solubility product of barium sulfate can be calculated by the square of the concentration of ions.
${{K}_{sp}}={{(M)}^{2}}$
${{K}_{sp}}={{(M)}^{2}}\text{= }{{\left( \dfrac{y\text{ x }{{10}^{3}}}{2({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})} \right)}^{2}}$
 ${{K}_{sp}}=\dfrac{{{10}^{6}}{{y}^{2}}}{4{{({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}^{2}}}$

Therefore, the correct answer is an option (b)- $\dfrac{{{10}^{6}}{{y}^{2}}}{4{{({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})}^{2}}}$

Note: We have subtracted twice the equivalent conductance of hydrochloric acid in the equation because 2 moles of hydrochloric acid were required to remove to get the equivalent conductance of barium sulfate.