If the lines frac{1-x}{3} = frac{7y-14}{2p} = | vedclass

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(D) First, rewrite the lines in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{x-1}{-3} = \frac{y-

Vedclass · Accepted Answer

$-\frac{70}{11}$

Vedclass · Answer

(D) First, rewrite the lines in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{-2}$. The direction vector is $\vec{a} = (-3, \frac{2p}{7}, -2)$.For the second line: $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$. The direction vector is $\vec{b} = (-\frac{3p}{7}, 1, -5)$.Since the lines are perpendicular, their dot product must be zero: $\vec{a} \cdot \vec{b} = 0$.$(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (-2)(-5) = 0$.$\frac{9p}{7} + \frac{2p}{7} + 10 = 0$.$\frac{11p}{7} = -10$.$p = -\frac{70}{11}$.

If the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{-2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular, then the value of $p$ is . . . . . . .

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