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3D Geometry question

2024 · 6 Apr · Shift 2 · Q54

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JEE Main
Mathematics
3D Geometry
2024 · 6 Apr · Shift 2 · Q54

3D Geometry question

2024 · 6 Apr · Shift 2 · Q54

JEE MainMathematics3D GeometryNumerical+4 / −1

If the shortest distance between the lines

\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}

and

\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}

is

\frac{44}{\sqrt{30}}

, then the largest possible value of

|\lambda|

is equal to

\underline{\hspace{2cm}}

.

Numerical answer

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