Verified Solution • 2025–26 Session moderate

QJEEM24Ma8de1: JEE Main 2024 Mathematics - Class 12 JEEM24Ma8de1 Solution: Let be the distance of the point of intersection of the lines…

Let d\mathrm{d} be the distance of the point of intersection of the lines x+63=y2=z+11\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1} and x−74=y−93=z−42\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2} from the point (7,8,9)(7,8,9) . Then d2+6\mathrm{d}^2+6 is equal to :

Multiple Choice Options

  • A 75
  • B 78
  • C 72
  • D 69

Step-by-Step Verified Solution

amp;P1:(3k−6,2k,k−1)amp;P2(4α+7,3α+9,2α+4)amp;P1≡P2amp;3k−6=4α+7⇒3k−4α=13amp;2k=3α+9⇒2k−3α=9amp;∴k=3,α=−1amp;∴P1:(3,6,2)\begin{aligned}& P_1:(3 k-6,2 k, k-1) \\& P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \\& P_1 \equiv P_2 \\& 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \\& 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \\& \therefore k=3, \alpha=-1 \\& \therefore P_1:(3,6,2)\end{aligned}

Distance of (3,6,2)(3,6,2) and (7,8,9)(7,8,9)

amp;=16+4+49=69=damp;d2+6=69+6=75\begin{aligned}& =\sqrt{16+4+49}=\sqrt{69}=d \\& d^2+6=69+6=75\end{aligned}

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