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(C) Given $|\vec{v} - \hat{i}| = |\vec{v} - 2\hat{j}| = |\vec{v} - \hat{j}|$.Let $\vec{v} = x\hat{i} + y\hat{j}$. The points are $A(1, 0)$,$B(0, 2)$,a

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$(2, 3]$

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(C) Given $|\vec{v} - \hat{i}| = |\vec{v} - 2\hat{j}| = |\vec{v} - \hat{j}|$.Let $\vec{v} = x\hat{i} + y\hat{j}$. The points are $A(1, 0)$,$B(0, 2)$,and $C(0, 1)$.Since $\vec{v}$ is equidistant from $A, B$,and $C$,it is the circumcenter of $	riangle ABC$.Equating the squared distances:$|\vec{v} - \hat{i}|^2 = |\vec{v} - \hat{j}|^2 \Rightarrow (x-1)^2 + y^2 = x^2 + (y-1)^2$$x^2 - 2x + 1 + y^2 = x^2 + y^2 - 2y + 1 \Rightarrow x = y$.Equating $|\vec{v} - \hat{j}|^2 = |\vec{v} - 2\hat{j}|^2$:$x^2 + (y-1)^2 = x^2 + (y-2)^2$$y^2 - 2y + 1 = y^2 - 4y + 4 \Rightarrow 2y = 3 \Rightarrow y = \frac{3}{2}$.Since $x = y$,we have $x = \frac{3}{2}$.Thus,$\vec{v} = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}$.$|\vec{v}| = \sqrt{(\frac{3}{2})^2 + (\frac{3}{2})^2} = \sqrt{\frac{9}{4} + \frac{9}{4}} = \sqrt{\frac{18}{4}} = \sqrt{4.5} \approx 2.12$.Since $2 < 2.12 \leq 3$,$|\vec{v}|$ lies in the interval $(2, 3]$.

Let $\vec{v}$ be a vector in the plane such that $|\vec{v} - \hat{i}| = |\vec{v} - 2\hat{j}| = |\vec{v} - \hat{j}|$. Then,$|\vec{v}|$ lies in the interval

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