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If $ \frac { 1 }{ 3 } (\frac { 1 }{ 4 } x-3)+2=\frac { 1 }{ 4 } x-\frac { 2 }{ 3 } (x-\frac { 3 }{ 2 } )+\frac { 1 }{ 2 } x $, then what is the value of $ x $?

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$0$

$12$

There is no solution for $ x $.

There are infinitely many solutions for $ x $.