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(D) First,find the unit vectors in the given directions:Let $\vec{a} = (2, -2, 1)$. The magnitude is $|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+

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$2\sqrt{10}$

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(D) First,find the unit vectors in the given directions:Let $\vec{a} = (2, -2, 1)$. The magnitude is $|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$.The unit vector is $\hat{a} = \frac{1}{3}(2, -2, 1)$.Since $\vec{x}$ has magnitude $6$,$\vec{x} = 6 \hat{a} = 2(2, -2, 1) = (4, -4, 2)$.Let $\vec{b} = (1, 1, -1)$. The magnitude is $|\vec{b}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$.The unit vector is $\hat{b} = \frac{1}{\sqrt{3}}(1, 1, -1)$.Since $\vec{y}$ has magnitude $\sqrt{3}$,$\vec{y} = \sqrt{3} \hat{b} = (1, 1, -1)$.Now,calculate $\vec{x} + 2\vec{y}$:$\vec{x} + 2\vec{y} = (4, -4, 2) + 2(1, 1, -1) = (4+2, -4+2, 2-2) = (6, -2, 0)$.Finally,find the magnitude $|\vec{x} + 2\vec{y}| = \sqrt{6^2 + (-2)^2 + 0^2} = \sqrt{36 + 4 + 0} = \sqrt{40} = 2\sqrt{10}$.

Vector $\vec{x}$ is a vector in the direction of $(2, -2, 1)$ and has a magnitude of $6$ units. Vector $\vec{y}$ is a vector in the direction of $(1, 1, -1)$ and has a magnitude of $\sqrt{3}$ units. Then,$|\vec{x} + 2\vec{y}| = $ . . . . . . .

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