(i) Let points (0,7,-10),(1,6,-6), and (4,9,-6) be denoted by A, B, and C respectively.AB=(1−0)2+(6−7)2+(−6+10)2​ =(1)2+(−1)2+(4)2​ =1+1+16​ =18​ =32​ BC=(4−1)2+(9−6)2+(−6+6)2​ =(3)2+(3)2​ =9+9​=18​=32​ CA=(0−4)2+(7−9)2+(−10+6)2​ =(−4)2+(−2)2+(−4)2​ =16+4+16​=36​=6 Here,A B = B C =CA Thus, the given points are the vertices of an isosceles triangle.(i) Let (0,7,10),(-1,6,6), and (-4,9,6) be denoted by A, B, and C respectively.=(−1)2+(−1)2+(−4)2​ =1+1+16​=18​ =32​ BC = (−4+1)2+(9−6)2+(6−6)2​ =(−3)2+(3)2+(0)2​ =9+9​=18​ =32​ CA=(0+4)2+(7−9)2+(10−6)2​ =(4)2+(−2)2+(4)2​ =16+4+16​ =36​ =6Now, AB2+BC2=(32​)2+(32​)2=18+18=36=AC2 Therefore, by Pythagoras theorem, ABC is a right triangle.Hence, the given points are the vertices of a right-angled triangle.(ii) Let (-1,2,1),(1,-2,5),(4,-7,8), and (2,-3,4) be denoted by A, B, C, and D respectively.AB =(1+1)2+(−2−2)2+(5−1)2​ =36​ =6BC =(4−1)2+(−7+2)2+(8−5)2​ =9+25+9​=43​ CD=(2−4)2+(−3+7)2+(4−8)2​ =4+16+16​ =36​ =6DA=(−1−2)2+(2+3)2+(1−4)2​ =9+25+9​=43​ Here, A B = C D = 6, B C = A D =43​ Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.Therefore, ABCD is a parallelogram.Hence, the given points are the vertices of a parallelogram.