If $\cos \alpha + \cos \beta + \cos \lambda = 0 = \sin \alpha + \sin \beta + \sin \gamma $ then which of the following is correct?
(A) ${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = \dfrac{3}{2}$
(B) $\sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3\sin \left( {\alpha + \beta + \gamma } \right)$
(C) $\cos \left( {\alpha + \beta } \right) + \cos \left( {\beta + \gamma } \right) + \cos \left( {\gamma + \alpha } \right) = 0$

Hint:
Use the given information to check for each of the three options. For option (A), use the cosine double angle formula to express it in terms of $2\alpha ,2\beta ,2\gamma $ and then evaluate the required value. For option (B), use identity triple angle formula for sine and then the identity ${x^3} + {y^3} + {z^3} = 3xyz$${\text{when }}x + y + z = 0$ to find right-hand side. For option (C), just use the addition formula for cosine and use previous relations to evaluate its value.

Complete step by step solution:
Here we are given two equations $\cos \alpha + \cos \beta + \cos \lambda = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$ . And with this information, we need to check whether the three given options are correct or not.
Now let’s rewrite the given equation as:
$\sin \alpha + \sin \beta = - \sin \gamma {\text{ }}...........(i)$ and $\cos \alpha + \cos \beta = - \cos \lambda {\text{ }}...........(ii)$
After squaring and adding both the equations, we get:
$ \Rightarrow {\left( {\sin \alpha + \sin \beta } \right)^2} + {\left( {\cos \alpha + \cos \beta } \right)^2} = {\sin ^2}\gamma + {\cos ^2}\gamma $
Using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and expand the above equation:
$ \Rightarrow {\sin ^2}\alpha + {\sin ^2}\beta + 2\sin \alpha \sin \beta + {\cos ^2}\alpha + {\cos ^2}\beta + 2\cos \alpha \cos \beta = {\sin ^2}\gamma + {\cos ^2}\gamma $
As we know the identity ${\sin ^2}x + {\cos ^2}x = 1$ , we can use it in the above equation as:
$ \Rightarrow 1 + 2\sin \alpha \sin \beta + 1 + 2\cos \alpha \cos \beta = 1 \Rightarrow 2 + 2\left( {\sin \alpha \sin \beta + \cos \alpha \cos \beta } \right) = 1$
Using the addition formula for cosine, i.e. $\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $ in the above equation, we get:
$ \Rightarrow 2 + 2\left( {\sin \alpha \sin \beta + \cos \alpha \cos \beta } \right) = 1 \Rightarrow 2 + 2\cos \left( {\alpha - \beta } \right) = 1$ ……….(iii)

Again squaring and subtracting (ii) from (i) as:
$ \Rightarrow {\left( {\cos \alpha + \cos \beta } \right)^2} - {\left( {\sin \alpha + \sin \beta } \right)^2} = {\cos ^2}\gamma - {\sin ^2}\gamma $
Expanding it similarly as previously using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ :
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + 2\cos \alpha \cos \beta - \left( {{{\sin }^2}\alpha + {{\sin }^2}\beta + 2\sin \alpha \sin \beta } \right) = {\cos ^2}\gamma - {\sin ^2}\gamma $
Now we can use the formula for cosine double angle, i.e. $\cos 2x = {\cos ^2}x - {\sin ^2}x$ and addition formula for cosine, i.e. $\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $ :
$ \Rightarrow \cos 2\alpha + \cos 2\beta + 2\cos \left( {\alpha + \beta } \right) = \cos 2\gamma $
Using the cosine addition formulas, which is: $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$ \Rightarrow \cos 2\alpha + \cos 2\beta + 2\cos \left( {\alpha + \beta } \right) = \cos 2\gamma \Rightarrow 2\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right) + 2\cos \left( {\alpha + \beta } \right) = \cos 2\gamma $
We can rearrange by taking common in the above equation as:
$ \Rightarrow \cos \left( {\alpha + \beta } \right)\left( {2\cos \left( {\alpha - \beta } \right) + 2} \right) = \cos 2\gamma $
Now using the relation (iii), we get:
$ \Rightarrow \cos \left( {\alpha + \beta } \right)\left( {2\cos \left( {\alpha - \beta } \right) + 2} \right) = \cos 2\gamma \Rightarrow \cos \left( {\alpha + \beta } \right) = \cos 2\gamma $ ……………..(iv)
From relation (iii), we say that:
$ \Rightarrow 2 + 2\cos \left( {\alpha - \beta } \right) = 1 \Rightarrow \cos \left( {\alpha - \beta } \right) = - \dfrac{1}{2}$ …………(v)
Now let’s consider option (A) and evaluate it using identity $2{\cos ^2}x = 1 + \cos 2x$ , we get:
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = \left( {\dfrac{{1 + \cos 2\alpha }}{2}} \right) + \left( {\dfrac{{1 + \cos 2\beta }}{2}} \right) + \left( {\dfrac{{1 + \cos 2\gamma }}{2}} \right)$
This can be further rearranged as:
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = \dfrac{3}{2} + \dfrac{1}{2}\left( {\cos 2\alpha + \cos 2\beta + \cos 2\gamma } \right)$
We can use $\cos 2\alpha + \cos 2\beta = 2\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right)$ in the above equation:
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = \dfrac{3}{2} + \dfrac{1}{2}\left( {\cos 2\alpha + \cos 2\beta + \cos 2\gamma } \right) = \dfrac{3}{2} + \dfrac{1}{2}\left( {2\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right) + \cos 2\gamma } \right)$
But from relation (v), we can substitute the value:
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = \dfrac{3}{2} + \dfrac{1}{2}\left( {2\cos \left( {\alpha + \beta } \right) \times \left( {\dfrac{{ - 1}}{2}} \right) + \cos 2\gamma } \right) = \dfrac{3}{2} + \dfrac{1}{2}\left( { - \cos \left( {\alpha + \beta } \right) + \cos 2\gamma } \right)$
From substituting relation (iv) we get:
\[ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = \dfrac{3}{2} + \dfrac{1}{2}\left( { - \cos \left( {\alpha + \beta } \right) + \cos 2\gamma } \right) = \dfrac{3}{2} + \dfrac{1}{2}\left( { - \cos 2\gamma + \cos 2\gamma } \right) = \dfrac{3}{2}\]
Therefore, we get: \[{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = \dfrac{3}{2}\]
Considering the option (B) and using the triple angle identity, i.e. $\sin 3x = 3\sin x - 4{\sin ^3}x$ , we get:
$ \Rightarrow \sin 3\alpha + \sin 3\beta + \sin 3\gamma = \left( {3\sin \alpha - 4{{\sin }^3}\alpha } \right) + \left( {3\sin \beta - 4{{\sin }^3}\beta } \right) + \left( {3\sin \gamma - 4{{\sin }^3}\gamma } \right)$
Now it can be rearranged as follows:
$ \Rightarrow \sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3\left( {\sin \alpha + \sin \beta + \sin \gamma } \right) - 4\left( {{{\sin }^3}\alpha + {{\sin }^3}\beta + {{\sin }^3}\gamma } \right)$
In the above expression, we can use (i) and the identity ${x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)$ ${\text{ when }}x + y + z = 0;{\text{ }}$ ${x^3} + {y^3} + {z^3} - 3xyz = 0 \Rightarrow {x^3} + {y^3} + {z^3} = 3xyz$ ................(vi)
Using the identity (vi) in the above equation where $\sin \alpha + \sin \beta + \sin \gamma = 0$ then:
$ \Rightarrow {\sin ^3}\alpha + {\sin ^3}\beta + {\sin ^3}\gamma = 3 \times \sin \alpha \times \sin \beta \times \sin \gamma $
Therefore,
$ \Rightarrow \sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3 \times 0 - 4\left( {3 \times \sin \alpha \times \sin \beta \times \sin \gamma } \right) = - 12\sin \alpha \sin \beta \sin \gamma $
Now considering the option (C), from relation (iv) we get $\cos \left( {\alpha + \beta } \right) = \cos 2\gamma $ and similarly we can conclude that $\cos \left( {\beta + \gamma } \right) = \cos 2\alpha $ and $\cos \left( {\gamma + \alpha } \right) = \cos 2\beta $
For option (C):
$ \Rightarrow \cos \left( {\alpha + \beta } \right) + \cos \left( {\beta + \gamma } \right) + \cos \left( {\gamma + \alpha } \right) = \cos 2\alpha + \cos 2\beta + \cos 2\gamma $
From the above equations, we concluded that $\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0$
Therefore, we get: $ \Rightarrow \cos \left( {\alpha + \beta } \right) + \cos \left( {\beta + \gamma } \right) + \cos \left( {\gamma + \alpha } \right) = \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0$
So, option (A) and (B) satisfies correctly but option (C) is not true

Hence, the option (A) and (B) are the correct answer.

Note:
In questions like this, do not forget to check for all of the options separately since it is a multiple-choice question. The use of different trigonometric formulas was the most crucial part of the solution. Go step by step and figure out ways to utilize the results from previous steps. Solve the left-side of all the options separately.