The vertices of the triangle are given to be $A(1,2,0)$, $B(0,0,0)$ and $C(-2,1,0)$. The question asks us whether this triangle is acute, obtuse or right.
Recall: A triangle is obtuse if any one angle of the triangle is obuse-angled. It is a right triangle if any one angle is a right angle. Finally, it is an acute triangle if all the three angles are acute. So to do the problem, we need to check whether each of the angles is acute or obtuse or right. I will show the steps for one angle. You can do for the other two angles similarly.
First of all, let $\theta$ be some angle in $(0, \pi)$. Then
- $\theta$ is acute $\iff$ $\theta < \frac{\pi}{2}$ $\iff$ $\cos \theta > 0$.
- $\theta$ is right $\iff$ $\theta = \frac{\pi}{2}$ $\iff$ $\cos \theta = 0$.
- $\theta$ is obtuse $\iff$ $\theta > \frac{\pi}{2}$ $\iff$ $\cos \theta < 0$.
Now, to see if (say) angle $A$ of the triangle $ABC$ is acute/right/obtuse, we need to check whether $\cos \angle BAC$ is positive/zero/negative. But what is $\cos \angle BAC$? It is the angle made by the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. (When you are computing the angle at a particular vertex $v$, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex $v$ as the initial point.) We will first compute these two vectors: $ \overrightarrow{AB} = (0,0,0) - (1,2,0) = (-1,-2,0) $ $ \overrightarrow{AC} = (-2,1,0) - (1,2,0) = (-3,-1,0) $ Therefore, the angle between these vectors is given by: $ \cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \ldots \tag{1} $ Can you take it from here? From the sign of this value, you should be able to decide if angle $A$ is acute/right/obtuse.
Now, do the same procedure for the remaining two angles $B$ and $C$ as well. That should help you solve the problem.
A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.