Let A and B denote the statements:
A: $\cos \alpha + \cos \beta + \cos \gamma = 0$
B: $\sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) = - \dfrac{3}{2}$
Then:
A) A is correct and B is incorrect
B) A is incorrect and B is correct
C) Both A and B are correct
D) Both A and B are incorrect

Hint: We are given two statements we have to find whether these statements hold when we are given the equation:
$\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) = - \dfrac{3}{2}$
We will first move the denominator of the fraction given on the right side to the left side and then square the whole equation and remember we will express $3$ as :
$3 = si{n^2}\alpha {\text{ }} + {\text{ }}co{s^2}\alpha {\text{ }} + {\text{ }}si{n^2}\beta {\text{ }} + {\text{ }}co{s^2}\beta {\text{ }} + {\text{ }}si{n^2}\gamma {\text{ }} + {\text{ }}co{s^2}\gamma {\text{ }}$ and then we apply the pythagorean trigonometric identity \[\;si{n^2}A{\text{ }} + {\text{ }}co{s^2}A{\text{ }} = {\text{ }}1\] to get desired answer.

Complete step by step answer:
We are given the two statements A and B and also the equation:
$\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) = - \dfrac{3}{2}$
\[2{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\gamma {\text{ }} - {\text{ }}\alpha } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\alpha {\text{ }} - {\text{ }}\beta } \right){\text{ }} = {\text{ }} - 3\]
Moving $3$ to the left hand side of the equation.
\[2{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\gamma {\text{ }} - {\text{ }}\alpha } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\alpha {\text{ }} - {\text{ }}\beta } \right){\text{ }} + {\text{ }}3{\text{ }} = {\text{ }}0\]
\[
  2{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\gamma {\text{ }} - {\text{ }}\alpha } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\alpha {\text{ }} - {\text{ }}\beta } \right){\text{ }} + {\text{ }}si{n^2}\alpha {\text{ }} + {\text{ }}co{s^2}\alpha {\text{ }} + {\text{ }}si{n^2}\beta {\text{ }} + {\text{ }}co{s^2}\beta {\text{ }} + \\
  {\text{ }}si{n^2}\gamma {\text{ }} + {\text{ }}co{s^2}\gamma {\text{ }} = {\text{ }}0 \\
\]
We will now solve the terms \[2{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\gamma {\text{ }} - {\text{ }}\alpha } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\alpha {\text{ }} - {\text{ }}\beta } \right){\text{ }}\]
We know the identity:
 \[cos\left( {A{\text{ }} - {\text{ }}B} \right){\text{ }} = {\text{ }}cos{\text{ }}A{\text{ }}cos{\text{ }}B{\text{ }} + {\text{ }}sin{\text{ }}A{\text{ }}sin{\text{ }}B\]
This will result in first term being written as : \[2{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ = }}2{\text{ }}cos\beta cos\gamma + 2{\text{ }}sin\beta sin\gamma \]
Other terms will also be similar and thus will result in the equation becoming
\[ 2{\text{ }}cos\beta cos\gamma + 2{\text{ }}sin\beta sin\gamma {\text{ }} + {\text{ }}2{\text{ }}cos\gamma cos\alpha + 2{\text{ }}sin\gamma sin\alpha {\text{ }} + {\text{ }}2{\text{ }}cos\alpha cos\beta + 2{\text{ }}sin\alpha sin\beta {\text{ }} + {\text{ }} \\
  si{n^2}\alpha {\text{ }} + {\text{ }}co{s^2}\alpha {\text{ }} + {\text{ }}si{n^2}\beta {\text{ }} + {\text{ }}co{s^2}\beta {\text{ }} + {\text{ }}si{n^2}\gamma {\text{ }} + {\text{ }}co{s^2}\gamma {\text{ }} = {\text{ }}0 \\
\]
We can now see that the left hand side is a perfect square identity of the form;
\[\;{\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)^2}\; = {\text{ }}{a^2}\; + {\text{ }}{b^2}\; + {\text{ }}{c^2}\; + {\text{ }}2\left( {ab{\text{ }} + {\text{ }}bc{\text{ }} + {\text{ }}ac} \right)\]
\[{\left( {cos{\text{ }}\alpha {\text{ }} + {\text{ }}cos{\text{ }}\beta {\text{ }} + {\text{ }}cos{\text{ }}\gamma } \right)^2}\; + {\text{ }}{\left( {sin{\text{ }}\alpha {\text{ }} + {\text{ }}sin{\text{ }}\beta {\text{ }} + {\text{ }}sin{\text{ }}\gamma } \right)^2}\; = {\text{ }}0\]
Since for this statement to be true both the expressions must be equal to zero and hence both the statements are true. The option C is hence true.

Note:
While applying the pythagorean trigonometric identity we should be careful regarding the angle. That is both the sine and cosine function angle should be the same. If we have \[{\sin ^2}x + {\cos ^2}y\] then it is not equal to 1, both the angles should be $x$ or it should be $y$. We used the same concept above.