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(D) Let the vertices of the triangle be $A = 2 \hat{i}+3 \hat{j}+5 \hat{k}$,$B = 5 \hat{i}+2 \hat{j}+3 \hat{k}$,and $C = 3 \hat{i}+5 \hat{j}+2 \hat{k}

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$x=y=z$

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(D) Let the vertices of the triangle be $A = 2 \hat{i}+3 \hat{j}+5 \hat{k}$,$B = 5 \hat{i}+2 \hat{j}+3 \hat{k}$,and $C = 3 \hat{i}+5 \hat{j}+2 \hat{k}$.Calculate the side lengths:$|AB| = |(5-2)\hat{i} + (2-3)\hat{j} + (3-5)\hat{k}| = |3\hat{i} - \hat{j} - 2\hat{k}| = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9+1+4} = \sqrt{14}$.$|BC| = |(3-5)\hat{i} + (5-2)\hat{j} + (2-3)\hat{k}| = |-2\hat{i} + 3\hat{j} - \hat{k}| = \sqrt{(-2)^2 + 3^2 + (-1)^2} = \sqrt{4+9+1} = \sqrt{14}$.$|AC| = |(3-2)\hat{i} + (5-3)\hat{j} + (2-5)\hat{k}| = |\hat{i} + 2\hat{j} - 3\hat{k}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$.Since $|AB| = |BC| = |AC| = \sqrt{14}$,the triangle is an equilateral triangle.For an equilateral triangle,the orthocentre coincides with the centroid.The centroid $G$ is given by $\frac{A+B+C}{3}$:$G = \frac{(2+5+3)\hat{i} + (3+2+5)\hat{j} + (5+3+2)\hat{k}}{3} = \frac{10\hat{i} + 10\hat{j} + 10\hat{k}}{3} = \frac{10}{3}\hat{i} + \frac{10}{3}\hat{j} + \frac{10}{3}\hat{k}$.Wait,re-calculating the sum: $(2+5+3) = 10$,$(3+2+5) = 10$,$(5+3+2) = 10$. The centroid is $\frac{10}{3}(\hat{i}+\hat{j}+\hat{k})$.Given the orthocentre is $x\hat{i}+y\hat{j}+z\hat{k}$,we have $x=y=z = \frac{10}{3}$.Thus,$x=y=z$ is the correct relation.

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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