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Maths Question 15 – JEE-MAIN 2026

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Let the image of the point P(1,6,a) in the line L:x1=y−12=z−a+1b, b>0, be (a3,0,a+c). If S(α,β,γ), α>0, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is 214, then α+β+γ is equal to:

The foot of the perpendicular from a point to a line is the midpoint of the segment connecting the point and its image in the line.

Step 1: Determine the foot of the perpendicular and line parameters✦ Active

Let F be the foot of the perpendicular from P(1,6,a) to the line L:x1=y−12=z−a+1b=λ. A general point on L is Q(λ,2λ+1,bλ+a−1). The vector PF→=(λ−1,2λ−5,bλ−1) must be perpendicular to the direction vector of L, d→=(1,2,b). Thus, PF→⋅d→=0, which gives (5+b2)λ−(11+b)=0. The image P′(a3,0,a+c) means F is the midpoint of PP′. So F=(1+a/32,6+02,a+a+c2)=(3+a6,3,2a+c2). Comparing the y-coordinate of F from both expressions: 2λ+1=3⟹λ=1. Substitute λ=1 into the dot product equation: (5+b2)(1)−(11+b)=0⟹b2−b−6=0⟹(b−3)(b+2)=0. Since b>0, b=3. Now, F=(1,2(1)+1,3(1)+a−1)=(1,3,a+2). Equating x and z coordinates of F: 3+a6=1⟹a=3. And 2a+c2=a+2⟹2a+c=2a+4⟹c=4. So, P(1,6,3), b=3, c=4. The line L is x1=y−12=z−23. The foot of the perpendicular F is (1,3,5).

Step 2: Find the coordinates of point S○ Expand

Point S(α,β,γ) is on line L, so S(λ,2λ+1,3λ+2). The distance FS=214. We use the distance formula:

FS2=(λ−1)2+(2λ+1−3)2+(3λ+2−5)2=(214)2=56

This simplifies to: (λ−1)2+(2λ−2)2+(3λ−3)2=56⟹(λ−1)2+4(λ−1)2+9(λ−1)2=56⟹14(λ−1)2=56⟹(λ−1)2=4. Thus, λ−1=±2, which gives λ=3 or λ=−1. For λ=3, S(3,2(3)+1,3(3)+2)=(3,7,11). Here α=3>0. For λ=−1, S(−1,2(−1)+1,3(−1)+2)=(−1,−1,−1). Here α=−1<0, so this solution is rejected. Thus, S(α,β,γ)=(3,7,11).

Step 3: Calculate α+β+γ○ Expand

From S(3,7,11), we have α=3,β=7,γ=11. The required sum is:

α+β+γ=3+7+11=21
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