Let d be the distance of the point of intersection of the lines (x+6)/3=y/2=(z+1)/1 and (x-7)/4=(y-9)/3=(z-4)/2 from the point (7,8,9). Then d²+6 is equal to:

Three Dimensional Geometry: JEE Main 2024 Mathematics question

JEE Main 2024 · 5 Apr, Morning shift · Q2539

JEE Main 20245 Apr, Morning shiftMathematicsMediumThree Dimensional Geometry249× since 2002

Let

\mathrm{d}

be the distance of 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}

from the point

(7,8,9)

. Then

\mathrm{d}^2+6

is equal to :