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$If \vec{x} and \vec{y} are two non-collinear vectors and A B C is a triangle with side lengths a, b and c satisfying$
$(20 a - 15b) \vec{x} + (15 b - 12 c) \vec{y} + (12 c - 20 a) (\vec{x} \times \vec{y}) = \vec{0}, then triangle ABC is$

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an obtuse-angled triangle

an acute-angled triangle

an isosceles triangle

a right-angled triangle

answer is C.

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Detailed Solution

$Since \vec{x}, \vec{y} and \vec{x} \times \vec{y} are linearly independent, we have$

$\begin{array}{l} 20 a - 15 b = 15 b - 12 c = 12 c - 20 a = 0 \\ \Rightarrow \frac{a}{3} = \frac{b}{4} = \frac{c}{5} \\ \Rightarrow c^{2} = a^{2} + b^{2} \end{array}$

Hence, $∆$ ABC is right angled.

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