If the orthocentre of the triangle whose vertices are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k},$ and $3 \hat{i} + 5 \hat{j} + 2 \hat{k}$
is $x \hat{i} + y \hat{j} + z \hat{k},$ then

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$x = y = z$

$x = 2 y = z$

$x = y = - z$

$x = y = 2 z$

answer is D.

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Detailed Solution

The vertices of triangle are
$2 \hat{i} + 3 \hat{j} + 5 \hat{k}$ , $5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $3 \hat{i} + 5 \hat{j} + 2 \hat{k}$
$A B = 3 \hat{i} - \hat{j} - 2 \hat{k} B C = - 2 \hat{i} + 3 \hat{j} - \hat{k} A C = \hat{i} + 2 \hat{j} - 3 \hat{k} | A B | = \sqrt{9 + 1 + 4} = \sqrt{14} | B C | = \sqrt{4 + 9 + 1} = \sqrt{14} | A C | = \sqrt{1 + 4 + 9} = \sqrt{14}$
$∵ Δ A B C$ is an equilateral triangle
$∴$ Orthocentre of triangle is coincide with centroid
$∴$ Orthocentre of $Δ A B C$ is
$= (\frac{2 \hat{i} + 3 \hat{j} + 5 \hat{k} + 5 \hat{i} + 2 \hat{j} + 3 \hat{k} + 3 \hat{i} + 5 \hat{j} + 2 \hat{k}}{3})$
$= (\frac{10 \hat{i} + 10 \hat{j} + 10 \hat{k}}{3}) = \frac{10}{3} \hat{i} + \frac{10}{3} \hat{j} + \frac{10}{3} \hat{k}$
$x = y = z = \frac{10}{3}$