Angles dilemma

$1)$ $\tan A + \tan B + \tan C = 0$

We know ,in a traingle $ABC$ ,
$\ \boxed{tan A + \tan B + \tan C = 1 - \tan A\tan B\tan C }$ $\Rightarrow 0 = 1 - \tan A\tan B\tan C \Rightarrow \tan A\tan B\tan C = 1$

.Such triangle is not possible as the angles wont add up to $180^\circ$

$2)$ Using Sine Rule ,we can see that the sides will be, $a =2k , b = 3k , c = k$ We can observe that , $\boxed { a + c = 2k + k = 3k = b }$

Such traingle is not possible as voilates the property of any traingle that sum of any two sides of a traingle is greater than the third side, i.e,

$\boxed{a + b > c, a + c > b , b + c > a}$

$3)$ As seen above, $(a + b) ^{2} > c^{2}$ . Hence, the equation will hold true for some values of $a, b, c.$

Divide both sides by $2$ , we get, $\dfrac {\sqrt{2}(\sin A + \cos A)}{2} = \dfrac {\sqrt{3}}{2}$

$\sin A \cos 45^\circ + \ cos A \sin 45^\circ = \dfrac {\sqrt{3}}{2}$ $\Rightarrow \sin(A + 45^\circ) = \sin(60^\circ) \Rightarrow A = 15^\circ$

$4)$ $\cos (A + B) = \cos A \cos B - \sin A \sin B = \dfrac {\sqrt{3}}{4} - \dfrac {\sqrt{3}}{4} = 0\Rightarrow A + B = 90^\circ$

$\cos(A - B) = \cos A \cos B + \sin A \sin B = \dfrac {\sqrt{3}}{4} -+\dfrac {\sqrt{3}}{4} = \dfrac {\sqrt{3}}{2} \Rightarrow A - B = 60^\circ$

$A= +60^\circ, -60^\circ, B = +30^\circ, -30^\circ \Rightarrow \sin A + \sin B = -\dfrac {\sqrt{3}}{2} + \dfrac {-1}{2} = -\dfrac {\sqrt{3} + 1}{2} \neq -\dfrac {\sqrt{3} + 1}{2\sqrt{2}}$

Hence, 1 , 2 and 4 are wrong,

$\boxed{ANSWER : 1||2||4 = 124}$