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(D) Let the position vectors of the points be $\vec{A} = 4\hat{i}+\hat{j}+3\hat{k}$,$\vec{B} = 6\hat{i}-2\hat{j}-3\hat{k}$,and $\vec{C} = \hat{i}-\hat

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an isosceles triangle

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(D) Let the position vectors of the points be $\vec{A} = 4\hat{i}+\hat{j}+3\hat{k}$,$\vec{B} = 6\hat{i}-2\hat{j}-3\hat{k}$,and $\vec{C} = \hat{i}-\hat{j}-3\hat{k}$.We calculate the lengths of the sides $AB$,$BC$,and $CA$:$AB = |\vec{B} - \vec{A}| = |(6-4)\hat{i} + (-2-1)\hat{j} + (-3-3)\hat{k}| = |2\hat{i} - 3\hat{j} - 6\hat{k}| = \sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.$BC = |\vec{C} - \vec{B}| = |(1-6)\hat{i} + (-1-(-2))\hat{j} + (-3-(-3))\hat{k}| = |-5\hat{i} + 1\hat{j} + 0\hat{k}| = \sqrt{(-5)^2 + 1^2 + 0^2} = \sqrt{25 + 1} = \sqrt{26}$.$CA = |\vec{A} - \vec{C}| = |(4-1)\hat{i} + (1-(-1))\hat{j} + (3-(-3))\hat{k}| = |3\hat{i} + 2\hat{j} + 6\hat{k}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.Since $AB = CA = 7$,the triangle is an isosceles triangle.

The points in space represented by the position vectors $A = 4\hat{i}+\hat{j}+3\hat{k}$,$B = 6\hat{i}-2\hat{j}-3\hat{k}$,and $C = \hat{i}-\hat{j}-3\hat{k}$ form:

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