The distance of the point (2, 4, 0) from the | vedclass

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(A) Let the first line be $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1} = \lambda$. Then any point on this line is $(3\lambda - 6, 2\lambda, \lambda -

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$3$ units

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(A) Let the first line be $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1} = \lambda$. Then any point on this line is $(3\lambda - 6, 2\lambda, \lambda - 1)$.Let the second line be $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2} = \mu$. Then any point on this line is $(4\mu + 7, 3\mu + 9, 2\mu + 4)$.For the point of intersection,we equate the coordinates:$3\lambda - 6 = 4\mu + 7 \implies 3\lambda - 4\mu = 13$ (Equation $1$)$2\lambda = 3\mu + 9 \implies 2\lambda - 3\mu = 9$ (Equation $2$)Solving these equations: Multiply Eq $1$ by $2$ and Eq $2$ by $3$:$6\lambda - 8\mu = 26$$6\lambda - 9\mu = 27$Subtracting gives $\mu = -1$. Substituting $\mu = -1$ into Eq $2$: $2\lambda - 3(-1) = 9 \implies 2\lambda = 6 \implies \lambda = 3$.Check with $z$-coordinates: $\lambda - 1 = 3 - 1 = 2$ and $2\mu + 4 = 2(-1) + 4 = 2$. Since they match,the intersection point is $(3(3) - 6, 2(3), 3 - 1) = (3, 6, 2)$.The distance between $(2, 4, 0)$ and $(3, 6, 2)$ is $\sqrt{(3-2)^2 + (6-4)^2 + (2-0)^2} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$ units.

The distance of the point $(2, 4, 0)$ from the point of intersection of the lines $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}$ and $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}$ is

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