If the line frac{2-x}{3}=frac{3y-2}{4lambda+1 | vedclass

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(D) The given lines are $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ and $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$.First,rewrite the lines in sta

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(D) The given lines are $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ and $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$.First,rewrite the lines in standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.For the first line: $\frac{x-2}{-3}=\frac{y-2/3}{(4\lambda+1)/3}=\frac{z-4}{-1}$. The direction vector is $\vec{v_1} = (-3, \frac{4\lambda+1}{3}, -1)$.For the second line: $\frac{x+3}{3\mu}=\frac{y-1/2}{-3}=\frac{z-5}{-7}$. The direction vector is $\vec{v_2} = (3\mu, -3, -7)$.Since the lines are perpendicular,their dot product is zero: $\vec{v_1} \cdot \vec{v_2} = 0$.$(-3)(3\mu) + (\frac{4\lambda+1}{3})(-3) + (-1)(-7) = 0$.$-9\mu - (4\lambda+1) + 7 = 0$.$-9\mu - 4\lambda - 1 + 7 = 0$.$-4\lambda - 9\mu + 6 = 0$.$4\lambda + 9\mu = 6$.

If the line $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$,then $4\lambda+9\mu$ is equal to :

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