Range of $|\vec a - \vec b|^2+|\vec b - \vec c|^2+|\vec c - \vec a|^2$

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In $\mathbf R^3$, let $\vec a$, $\vec b$ and $\vec c$ be three vectors such that $|\vec a|=2$, $|\vec b|=3$ and $|\vec c|=4$. Then find the minimum and maximum possible value of the expression

$$|\vec a - \vec b|^2+|\vec b - \vec c|^2+|\vec c - \vec a|^2$$

I have tried the following:

|a-b|^2 + |b-c|^2 + |c-a|^2 = 2(a^2 + b^2 + c^2) -2(a.b + b.c + c.a) = 58 - 2(a.b + b.c + c.a)

Now, (a+b+c)^2 >= 0 => a^2 + b^2 + c^2 >= -2(a.b + b.c + c.a) => -2(a.b + b.c + c.a) <= 29 => |a-b|^2 + |b-c|^2 + |c-a|^2 <= 87

But I can't find the maximum value.

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Let $\measuredangle(\vec{c},\vec{b})=\alpha$, $\measuredangle(\vec{a},\vec{c})=\beta$ and $\measuredangle(\vec{a},\vec{b})=\gamma$. Thus, $$|\vec a - \vec b|^2+|\vec b - \vec c|^2+|\vec c - \vec a|^2=58-24\cos\alpha-16\cos\beta-12\cos\gamma,$$ which gives the minimum easily: $$58-24-16-12=6.$$

The maximum is $87$ because $$\sum_{cyc}(\vec{a}-\vec{b})^2=3(\vec{a}^2+\vec{b}^2+\vec{c}^2)-(\vec{a}+\vec{b}+\vec{c})^2\leq3(\vec{a}^2+\vec{b}^2+\vec{c}^2)=87$$