If frac{x}{cos alpha} = frac{y}{cos left(frac | vedclass

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(B) Let $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)} = k$. The

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(B) Let $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)} = k$. Then,$x = k \cos \alpha$,$y = k \cos \left(\frac{2 \pi}{3} - \alpha\right)$,and $z = k \cos \left(\frac{2 \pi}{3} + \alpha\right)$. Now,$x + y + z = k \left[ \cos \alpha + \cos \left(\frac{2 \pi}{3} - \alpha\right) + \cos \left(\frac{2 \pi}{3} + \alpha\right) \right]$. Using the formula $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$,we get: $x + y + z = k \left[ \cos \alpha + 2 \cos \left(\frac{2 \pi}{3}\right) \cos \alpha \right]$. Since $\cos \left(\frac{2 \pi}{3}\right) = -\frac{1}{2}$,we have: $x + y + z = k \left[ \cos \alpha + 2 \left(-\frac{1}{2}\right) \cos \alpha \right] = k [\cos \alpha - \cos \alpha] = 0$.

If $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)}$,then the value of $(x + y + z)$ is equal to

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