Question

show that the three points (-2,3,5),(1,2,3),(7,0,-1) are collinear.

Answer 1

Cal the area if it comes 0 then they r collinear

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

${\boxed{\sf\:{Points\;be\;ABC\;are\;collinear\;on\;a\;line }}}$

AB = √{(1 + 2)² + (2 - 3)² + (3 - 5)²}

AB = √{(3)² + (-1)² + (-2)²}

AB = √(9 + 1 + 4)

AB = √14

$\textbf{\underline{Here\;we\;get:-}}$

${\boxed{\sf\:{AB =\sqrt{14}}}}$

Now,

BC = √{(7 - 1)² + (0 - 2)² + (-1 - 3)²}

BC = √{(6)² + (-2)² + (-4)²}

BC = √(36 + 4 + 16)

BC = √(40 + 16)

BC = √56

${\boxed{\sf\:{BC=2\sqrt{14}}}}$

$\textbf{\underline{Here\;we\;get:-}}$

BC = 2√14

Also,

AC = √{(7 + 2)² + (0 - 3)² + (-1 - 5)²}

AC = √{(9)² + (-3)² + (-6)²}

AC = √(81 + 9 + 36)

AC = √(90 + 36)

AC = √120

${\boxed{\sf\:{AC=3\sqrt{14}}}}$

$\textbf{\underline{Here\;we\;get:-}}$

${\boxed{\sf\:{AC=3\sqrt{14}}}}$

Therefore,

AB + BC = AC

√14 + 2√14 = 3√14

3√14 = 3√14

Hence it is proved that points A(-2,3,5) , B(1,2,3) and C(7,0,-1) are collinear