Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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 :
MathematicsThree Dimensional GeometryJEE MainJEE Main 2024 (05 Apr Shift 1)
Options:
  • A 69
  • B 78
  • C 72
  • D 75
Solution:
2409 Upvotes Verified Answer
The correct answer is: 75
$\begin{aligned} & \frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda....(1)\\ & x=3 \lambda-6, y=2 \lambda, z=\lambda-1 \\ & \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu....(2)\end{aligned}$
$\mathrm{x}=4 \mu+7, \mathrm{y}=3 \mu+9, \mathrm{z}=2 \mu+4$
$3 \lambda-6=4 \mu+7 \Rightarrow 3 \lambda-4 \mu=13$ $\ldots(3) \times 2$
$2 \lambda=3 \mu+9 \Rightarrow 2 \lambda-3 \mu=9$ $\ldots(4) \times 3$
$\begin{array}{c}
6 \lambda-8 \mu=26 \\
6 \lambda-9 \mu=27 \\
-\quad+\quad- \\
\hline\mu=-1
\end{array}$
$\begin{gathered}
\Rightarrow 3 \lambda-4(-1)=13 \\
3 \lambda=9 \\
\lambda=3
\end{gathered}$
int. point $(3,6,2) ;(7,8,9)$
$\mathrm{d}^2=16+4+49=69$
Ans. $\mathrm{d}^2+6=69+6=75$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.