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(D) Given $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Since $\vec{a} \perp \vec{b}$, $\vec{a} \cdot \vec{b} = 0$.Since the angle between $\vec{b}$ and $\

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$(\vec{a} - \vec{c}) \cdot (\vec{b} - \vec{c}) = 2$

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(D) Given $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Since $\vec{a} \perp \vec{b}$, $\vec{a} \cdot \vec{b} = 0$.Since the angle between $\vec{b}$ and $\vec{c}$ is $120^\circ$, $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos(120^\circ) = 1 \cdot 1 \cdot (-1/2) = -1/2$.Given $(\vec{a} + \vec{c}) \perp (\vec{b} + \vec{c})$, so $(\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 0$.Expanding this: $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{c} \cdot \vec{b} + |\vec{c}|^2 = 0$.Substituting values: $0 + \vec{a} \cdot \vec{c} - 1/2 + 1 = 0$, which gives $\vec{a} \cdot \vec{c} = -1/2$.Now evaluate $(\vec{a} - \vec{c}) \cdot (\vec{b} - \vec{c}) = \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{b} + |\vec{c}|^2$.$= 0 - (-1/2) - (-1/2) + 1 = 1/2 + 1/2 + 1 = 2$.

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $120^\circ$. If $\vec{a} + \vec{c}$ is perpendicular to $\vec{b} + \vec{c}$, then:

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