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(C) The formula for the cosine of the angle $	heta$ between two lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ is given by $\cos 	h

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$\pm 5$

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(C) The formula for the cosine of the angle $	heta$ between two lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ is given by $\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$.Given $	heta = \frac{\pi}{3}$,$a_1=4, b_1=-3, c_1=5$ and $a_2=3, b_2=4, c_2=k$.Substituting these values,we get $\cos \frac{\pi}{3} = \left| \frac{4(3) + (-3)(4) + 5k}{\sqrt{4^2 + (-3)^2 + 5^2} \sqrt{3^2 + 4^2 + k^2}} ight|$.$\frac{1}{2} = \left| \frac{12 - 12 + 5k}{\sqrt{16 + 9 + 25} \sqrt{9 + 16 + k^2}} ight|$.$\frac{1}{2} = \left| \frac{5k}{\sqrt{50} \sqrt{25 + k^2}} ight| = \left| \frac{5k}{5\sqrt{2} \sqrt{25 + k^2}} ight| = \left| \frac{k}{\sqrt{2} \sqrt{25 + k^2}} ight|$.Squaring both sides,$\frac{1}{4} = \frac{k^2}{2(25 + k^2)}$.$2(25 + k^2) = 4k^2$.$50 + 2k^2 = 4k^2$.$2k^2 = 50 \Rightarrow k^2 = 25$.Therefore,$k = \pm 5$.

If the angle between the lines whose direction ratios are $4, -3, 5$ and $3, 4, k$ is $\frac{\pi}{3}$,then $k=$

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