Consider the line L passing through the point | vedclass

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(A) The equation of the line $L$ passing through $(1, 2, 3)$ and $(2, 3, 5)$ is given by:$\frac{x-1}{2-1} = \frac{y-2}{3-2} = \frac{z-3}{5-3} \Rightar

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$3$

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(A) The equation of the line $L$ passing through $(1, 2, 3)$ and $(2, 3, 5)$ is given by:$\frac{x-1}{2-1} = \frac{y-2}{3-2} = \frac{z-3}{5-3} \Rightarrow \frac{x-1}{1} = \frac{y-2}{1} = \frac{z-3}{2} = \lambda$Any point $B$ on line $L$ is $(1+\lambda, 2+\lambda, 3+2\lambda)$.The given line along which the distance is measured is $\frac{x-11/3}{2/3} = \frac{y-11/3}{1/3} = \frac{z-19/3}{2/3}$.This line passes through $A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ and has direction ratios $\langle 2, 1, 2 \rangle$.Since $B$ lies on this line,the vector $\vec{AB}$ must be parallel to $\langle 2, 1, 2 \rangle$.$\vec{AB} = \left(1+\lambda-\frac{11}{3}, 2+\lambda-\frac{11}{3}, 3+2\lambda-\frac{19}{3}\right) = \left(\lambda-\frac{8}{3}, \lambda-\frac{5}{3}, 2\lambda-\frac{10}{3}\right) = \frac{1}{3} \langle 3\lambda-8, 3\lambda-5, 6\lambda-10 \rangle$.Since $\vec{AB}$ is parallel to $\langle 2, 1, 2 \rangle$,we have $\frac{3\lambda-8}{2} = \frac{3\lambda-5}{1} = \frac{6\lambda-10}{2}$.From $\frac{3\lambda-8}{2} = 3\lambda-5$,we get $3\lambda-8 = 6\lambda-10 \Rightarrow 3\lambda = 2 \Rightarrow \lambda = \frac{2}{3}$.Substituting $\lambda = \frac{2}{3}$ in $\vec{AB}$,we get $\vec{AB} = \frac{1}{3} \langle 3(\frac{2}{3})-8, 3(\frac{2}{3})-5, 6(\frac{2}{3})-10 \rangle = \frac{1}{3} \langle -6, -3, -6 \rangle = \langle -2, -1, -2 \rangle$.The distance $AB = |\vec{AB}| = \sqrt{(-2)^2 + (-1)^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$. The distance of the point $A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $L$ along the line $\frac{3x-11}{2} = \frac{3y-11}{1} = \frac{3z-19}{2}$ is equal to:

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