If d₁ is the shortest distance between the lines x+1=2 y=-12 z, x=y+2=6 z-6 and d₂ is the shortest distance between the lines (x-1)/2=(y+8)/(-7)=(z-4)/5, (x-1)/2=(y-2)/1=(z-6)/(-3), then the value of (32 √(3) ~d₁)/(~d₂) is:

Three Dimensional Geometry: JEE Main 2024 Mathematics question

JEE Main 2024 · 30 Jan, Morning shift · Q24

JEE Main 202430 Jan, Morning shiftMathematicsMediumThree Dimensional Geometry249× since 2002

\mathrm{d}_1

is the shortest distance between the lines

x+1=2 y=-12 z, x=y+2=6 z-6

and

\mathrm{d}_2

is the shortest distance between the lines

\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}

, then the value of

\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}

is :

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2024 · 30 Jan, Morning shift · Q24

JEE Main 202430 Jan, Morning shiftMathematicsMediumThree Dimensional Geometry249× since 2002

\mathrm{d}_1

is the shortest distance between the lines

x+1=2 y=-12 z, x=y+2=6 z-6

and

\mathrm{d}_2

is the shortest distance between the lines

\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}

, then the value of

\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}

is :

Your answerType your answer. Integers and decimals are accepted.