Let the line x1=6-y2=z+85 intersect the lines x-54=y-73=z+21 and x+36=3-y3=z-61 at the poi…

Question

Let the line $\dfrac{x}{1}=\dfrac{6-y}{2}=\dfrac{z+8}{5}$ intersect the lines $\dfrac{x-5}{4}=\dfrac{y-7}{3}=\dfrac{z+2}{1}$ and $\dfrac{x+3}{6}=\dfrac{3-y}{3}=\dfrac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the midpoint of the line segment $AB$ from the plane $2x-2y+z=14$ is:

Accepted Answer

Writing the first line as $(\lambda,\ -2\lambda+6,\ 5\lambda-8)$:

Intersection with the second line gives $\lambda=1$: $A=(1,4,-3)$.
Intersection with the third line gives $\lambda=3$: $B=(3,0,7)$.

Midpoint $=(2,2,2)$.
$$\text{Distance}=\frac{|2(2)-2(2)+2-14|}{\sqrt{4+4+1}}=\frac{12}{3}=4.$$

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