If the vertices of a triangle $\mathrm{ABC}$ are $\mathrm{A}(1,2,3) \mathrm{B}(\mathrm{h},-3,0)$ and…
If the vertices of a triangle $\mathrm{ABC}$ are $\mathrm{A}(1,2,3) \mathrm{B}(\mathrm{h},-3,0)$ and $\mathrm{C}(-4, \mathrm{k},-1)$ and the centroid of the triangle is $\left(5,-1, \frac{2}{3}\right)$ then triangle $\mathrm{ABC}$ is
an obtuse angled triangle
an acute angled triangle
an isosceles triangle
a right angled triangle
Solution
Given vertices of a tiangle $\mathrm{ABC}$ are $\mathrm{A}(1,2,3), \mathrm{B}=$ $(\mathrm{h},-3,0)$ and $\mathrm{C}(-4, \mathrm{~K},-1)$ and controid of $\varnothing \mathrm{ABC}$ is $\left(5,-1, \frac{2}{3}\right)$ now $\frac{1+\mathrm{h}-4}{3}=5 \Rightarrow \mathrm{h}=18$ and $\frac{2+\mathrm{K}-3}{3}=-1 \Rightarrow \mathrm{k}=-2$
now $\mathrm{A}=(1,2,3) \mathrm{B}=(18,-3,0)$ and $\mathrm{G}=\left(5,-1, \frac{2}{3}\right)$ now $\mathrm{AB}=\sqrt{17^2+(-5)^2+(-3)^2}=\sqrt{323}$
$
\begin{aligned}
& \mathrm{BC}=\sqrt{(-22)^2+1^2+1^2}=\sqrt{486} \\
& \mathrm{CA}=\sqrt{(-5)^2+(-4)^2+(-4)^2}=\sqrt{57}
\end{aligned}
$
here $\mathrm{BC}^2>\mathrm{AB}^2+\mathrm{CA}^2$
$\Rightarrow$ Given triangle $\mathrm{ABC}$ is an obtuse triangle