Let the shortest distance between the lines (x-3)/3 = (y-α)/(-1) = (z-3)/1 and (x+3)/(-3) = (y+7)/2 = (z-β)/4 be 3√(30). Then the positive value of 5α + β is:

Three Dimensional Geometry: JEE Main 2025 Mathematics question

JEE Main 2025 · April, Unspecified (session-level dataset) · Q131

JEE Main 2025April, Unspecified (session-level dataset)MathematicsMediumThree Dimensional Geometry249× since 2002

Let the shortest distance between the lines

\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}

and

\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}

be 3\sqrt{30}. Then the positive value of 5\alpha + \beta is:

JEE Main 2025 · April, Unspecified (session-level dataset) · Q131

JEE Main 2025April, Unspecified (session-level dataset)MathematicsMediumThree Dimensional Geometry249× since 2002

Let the shortest distance between the lines

\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}

and

\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}

be 3\sqrt{30}. Then the positive value of 5\alpha + \beta is: