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The lines $2 \mathrm{x}=3 \mathrm{y}=-\mathrm{z}$ and $6 \mathrm{x}=-\mathrm{y}=-4 \mathrm{z}$
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are perpendicular
$2 \mathrm{x}=3 \mathrm{y}=-\mathrm{z}$
or $\frac{\mathrm{x}}{3}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{-6}$
$6 \mathrm{x}=-\mathrm{y}=-4 \mathrm{z}$
or $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{-12}=\frac{\mathrm{z}}{-3}$
$\cos \theta=\frac{\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}+\mathrm{z}_{1} \mathrm{z}_{2}}{\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{z}_{1}^{2}} \cdot \sqrt{\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}+\mathrm{z}_{3}^{2}}}$
$=\frac{(6-24+18)}{\sqrt{3^{2}+2^{2}+(-6)^{2}} \cdot \sqrt{2^{2}+(-12)^{2}+(-3)^{2}}}$
$\cos \theta=0$
$\theta=90^{\circ}$
So lines are perpendicular
or $\frac{\mathrm{x}}{3}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{-6}$
$6 \mathrm{x}=-\mathrm{y}=-4 \mathrm{z}$
or $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{-12}=\frac{\mathrm{z}}{-3}$
$\cos \theta=\frac{\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}+\mathrm{z}_{1} \mathrm{z}_{2}}{\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{z}_{1}^{2}} \cdot \sqrt{\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}+\mathrm{z}_{3}^{2}}}$
$=\frac{(6-24+18)}{\sqrt{3^{2}+2^{2}+(-6)^{2}} \cdot \sqrt{2^{2}+(-12)^{2}+(-3)^{2}}}$
$\cos \theta=0$
$\theta=90^{\circ}$
So lines are perpendicular
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