Answer:
Yes, it is right-angled
Step-by-step explanation:
Two vectors are orthogonal if the scalar product between them is zero. Then, we will match each pair of vertices with a vector, wich is formed with the following formula:
Given two points A and B, the vector going from A to B is
![AB=B-A=(B_(x)-A_(x),B_(y)-A_(y),B_(z) -A_(z))](https://img.qammunity.org/2020/formulas/physics/high-school/3ggwcvsirm07edz3wlv6lc41g82rxmrixp.png)
So, we calculate each component separately.
![PQ=Q-P=(2-1,1-(-2),-3-(-1))=(1,3,-2)](https://img.qammunity.org/2020/formulas/physics/high-school/n73kxuuumg5moekhkc8oe3zaokhevof2mb.png)
![QR=R-Q=(6-2,-1-1,-4-(-3))=(4,-2,-1)](https://img.qammunity.org/2020/formulas/physics/high-school/o7f3gvk3gy0cx83pv2p0tmc4yem0lj9pb4.png)
![RP=P-R=(1-6,-2-(-1),-1-(-4))=(-5,-1,3)](https://img.qammunity.org/2020/formulas/physics/high-school/qm1jnur9cvmpkv62nmqom5ygjq6ifrhp8p.png)
Finally, using the scalar product formula
![A*B=A_(x)* B_(x)+ A_(y)* B_(y)+ A_(z)* B_(z)](https://img.qammunity.org/2020/formulas/physics/high-school/1ey6druev78kyan4klo5zsgwdh4y2dse8q.png)
we see if the products is zero
![PQ*QR=1*4+3(-2)+(-2)*(-1)=0](https://img.qammunity.org/2020/formulas/physics/high-school/5zfaymlz7lmcudb833d8dva68xaddkbgbu.png)
In this case we don't even have to calculate the other products as we've found that PQ and QR form a right angle.