Match the following
I. The centroid of the triangle formed by $(2, 3, - 1), (5, 6, 3), (2, - 3, 1)$ is a) $(2, 2, 2)$
II. The circumcentre of the triangle formed by $(1, 2, 3), (2, 3, 1), (3, 1, 2)$ is b) $(3, 1, 4)$
III. The orthocenter of the triangle formed by $(2, 1, 5), (3, 2, 3), (4, 0, 4)$ is c) $(1, 1, 0)$
IV. The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 4, 0)$ is d) $(3, 2, 1)$
e) $(0, 0, 0)$

I - d, I I - a, I I I - b, I V - c

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I - a, I I - b, I I I - c, I V - d

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I - d, I I - e, I I I - b, I V - c

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I - d, I I - a, I I I - e, I V - c

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Solution

The correct option is A $I - d, I I - a, I I I - b, I V - c$
$I) centroid$ $= (\frac{2 + 5 + 2}{3}, \frac{3 + 6 - 3}{3}, \frac{- 1 + 3 + 1}{3})$
$= (3, 2, 1)$

$II) It is an equilateral triangle as its circumcenter will be its centroid. and centroid is$
$(\frac{1 + 2 + 3}{3}, \frac{2 + 3 + 1}{3}, \frac{3 + 1 + 2}{3}) = (2, 2, 2)$

$III)$ Let, P(x, y, z) be the orthocenter. So, $A P ⊥ B C, B P ⊥ A C$ and $C P ⊥ A B$ .
$D R^{s}$ of $A P = (x - 2, y - 1, z - 5)$
$D R^{s}$ of $B C = (1, - 2, 1)$
As, $A P ⊥ B C$ , $x - 2 - 2 y + 2 + z - 5 = 0$
$\Rightarrow x - 2 y + z = 5$
check all the remaining options.
Only (3, 1, 4) satisfies it.
So, (3, 1, 4) will be orthocenter

$IV) Incentre$
$a = \sqrt{9 + 16 + 0} = 5$
$b = 4, c = 3$
Incenter $= (\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}, \frac{a z_{1} + a z_{2} + a z_{3}}{a + b + c})$
$= (\frac{5 \times 0 + 4 \times 3 + 3 \times 0}{3 + 4 + 5}, \frac{5.0 + 4.0 + 3.4}{3 + 4 + 5}, \frac{5.0 + 4.0 + 3.0}{3 + 4 + 5})$
$= (1, 1, 0)$

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I. The centroid of the triangle formed by $(2, 3, - 1), (5, 6, 3), (2, - 3, 1)$ is	a) $(2, 2, 2)$
II. The circumcentre of the triangle formed by $(1, 2, 3), (2, 3, 1), (3, 1, 2)$ is	b) $(3, 1, 4)$
III. The orthocenter of the triangle formed by $(2, 1, 5), (3, 2, 3), (4, 0, 4)$ is	c) $(1, 1, 0)$
IV. The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 4, 0)$ is	d) $(3, 2, 1)$
	e) $(0, 0, 0)$