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(C) Given $a + b + c = 0$,we can write $a + b = -c$.Squaring both sides,we get $|a + b|^2 = |-c|^2$.Using the property $|x|^2 = x \cdot x$,we have $|a

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$60$

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(C) Given $a + b + c = 0$,we can write $a + b = -c$.Squaring both sides,we get $|a + b|^2 = |-c|^2$.Using the property $|x|^2 = x \cdot x$,we have $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.Since $a \cdot b = |a||b| \cos 	heta$,where $	heta$ is the angle between $a$ and $b$,the equation becomes $|a|^2 + |b|^2 + 2|a||b| \cos 	heta = |c|^2$.Substituting the given values: $3^2 + 5^2 + 2(3)(5) \cos 	heta = 7^2$.$9 + 25 + 30 \cos 	heta = 49$.$34 + 30 \cos 	heta = 49$.$30 \cos 	heta = 49 - 34 = 15$.$\cos 	heta = \frac{15}{30} = \frac{1}{2}$.Therefore,$	heta = 60^\circ$.

If three vectors $a, b, c$ satisfy $a + b + c = 0$ and $|a| = 3, |b| = 5, |c| = 7$,then the angle between $a$ and $b$ is ............ $^\circ$.

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