JEE Main 2024 — Mathematics Vectors & 3D Geometry

The given point is $P(3,4,9)$P(3,4,9).

Let, $Q(\alpha ,\beta ,\gamma )$Q(α,β,γ) be the image of $P(3,4,9)$P(3,4,9) and $'O'$′O′ be the mid-point of $PQ$PQ which is on the given line

$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}=\lambda$3x−1​=2y+1​=1z−2​=λ (say)

$$\Rightarrow x=3\lambda +1,y=2\lambda -1,z=\lambda +2$$⇒x=3λ+1,y=2λ−1,z=λ+2

let $(3\lambda +1,2\lambda -1,\lambda +2)$(3λ+1,2λ−1,λ+2) be the co-ordinates of $O$O.

Direction ratios of the line is ${a}_{1}=3,{b}_{1}=2,{c}_{1}=1$a1​=3,b1​=2,c1​=1

Also, direction ratios of $PO$PO are

$$(3\lambda -2,2\lambda -5,\lambda -7)$$(3λ−2,2λ−5,λ−7)

As, $PO$PO is perpendicular to the line

$$3(3\lambda -2)+2(2\lambda -5)+\lambda -7=0$$3(3λ−2)+2(2λ−5)+λ−7=0

$$\Rightarrow 14\lambda -23=0$$⇒14λ−23=0

$$\Rightarrow \lambda =\frac{23}{14}$$⇒λ=1423​

$\therefore$∴ $O$O point is $(3\cdot \frac{23}{14}+1,2\cdot \frac{23}{14}-1,\frac{23}{14}+2)$(3⋅1423​+1,2⋅1423​−1,1423​+2)

Since, $O$O is mid-point of $PQ$PQ

$$\therefore (3\cdot \frac{23}{14}+1,2\cdot \frac{23}{14}-1,\frac{23}{14}+2)=(\frac{3+\alpha }{2},\frac{4+\beta }{2},\frac{9+\gamma }{2})$$∴(3⋅1423​+1,2⋅1423​−1,1423​+2)=(23+α​,24+β​,29+γ​)

So, on comparing both side and solving we get,

$$\alpha =\frac{62}{7},\beta =\frac{4}{7},\gamma =\frac{-12}{7}$$α=762​,β=74​,γ=7−12​

Hence, $14(\alpha +\beta +\gamma )=108$14(α+β+γ)=108