Distance AB:
A=(1, 5, 0), B=(3, 8, 6)
AB = \(\sqrt {{{(3 - 1)}^2} + {{\left( {8 - 5} \right)}^2} + {{\left( { 6 -0} \right)}^2}} \)
=
7units
Distance BC:
B=(3, 8, 6) ,C=(7, −7, 4)
BC = \(\sqrt {{{(7 - 3)}^2} + {{\left( {-7 - 8} \right)}^2} + {{\left( { 4 -6} \right)}^2}} \)
=\(7\sqrt 5 \) units
Distance AC:
A= (1, 5, 0),C=(7, −7, 4)
AC = \(\sqrt {{{(7 - 1)}^2} + {{\left( {-7 - 5} \right)}^2} + {{\left( { 4 -0} \right)}^2}} \)
= 14 units
From the distances, we observe that
\(BC^2=AB^2+AC^2\)
\({\left( {7\sqrt 5 } \right)^2} = {7^2} + {14^2}\)
245 = 245
Thus, the given triangle is a right-angled triangle.