Vectors & 3D GeometrymediumPYQ · JEE Main · 4 Jun 2026 · Shift 1 (Morning)Free

Reflection in a Line: α+β+γ = 21 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let the image of the point P(1,6,a)P(1,6,a) in the line L:x1=y−12=z−a+1bL:\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-a+1}{b}, b>0b>0, be (a3,0,a+c)\left(\dfrac{a}{3},0,a+c\right). If S(α,β,γ)S(\alpha,\beta,\gamma), α>0\alpha>0, is the point on LL such that the distance of SS from the foot of perpendicular from PP on LL is 2142\sqrt{14}, then α+β+γ\alpha+\beta+\gamma is equal to
A1919
B2020
C2121correct
D2222
Solution
Step 1: On L:x1=y−12=z−a+1b=tL:\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-a+1}{b}=t, point (t, 1+2t, a−1+bt)(t,\ 1+2t,\ a-1+bt), direction d⃗=(1,2,b)\vec d=(1,2,b). Step 2: Q=Q= midpoint of P(1,6,a)P(1,6,a) and image (a3,0,a+c)\left(\dfrac{a}{3},0,a+c\right):
Q=(1+a/32, 3, 2a+c2).Q=\left(\dfrac{1+a/3}{2},\ 3,\ \dfrac{2a+c}{2}\right).
 Q∈LQ\in L, yy: 1+2t=3⇒t=11+2t=3\Rightarrow t=1; xx: t=1⇒1+a/32=1⇒1+a3=2⇒a=3t=1\Rightarrow\dfrac{1+a/3}{2}=1\Rightarrow1+\dfrac a3=2\Rightarrow a=3. Step 3: a=3,t=1a=3,t=1: Qz=a−1+b=2+b=6+c2Q_z=a-1+b=2+b=\dfrac{6+c}{2}. PQ⃗=(0,−3, Qz−3)⊥d⃗=(1,2,b)\vec{PQ}=(0,-3,\ Q_z-3)\perp\vec d=(1,2,b):
0−6+b(Qz−3)=0⇒b(Qz−3)=6.0-6+b(Q_z-3)=0\Rightarrow b(Q_z-3)=6.
Qz=2+b⇒b(b−1)=6⇒b2−b−6=0⇒(b−3)(b+2)=0.Q_z=2+b\Rightarrow b(b-1)=6\Rightarrow b^2-b-6=0\Rightarrow(b-3)(b+2)=0.
 b>0⇒b=3⇒Qz=5⇒6+c2=5⇒c=4b>0\Rightarrow b=3\Rightarrow Q_z=5\Rightarrow\dfrac{6+c}{2}=5\Rightarrow c=4. ∴ Q=(1,3,5)Q=(1,3,5). Step 4: b=3b=3: S(k, 2k+1, 3k+2)S(k,\ 2k+1,\ 3k+2).
SQ2=(k−1)2+(2k−2)2+(3k−3)2=(k−1)2(1+4+9)=14(k−1)2.SQ^2=(k-1)^2+(2k-2)^2+(3k-3)^2=(k-1)^2(1+4+9)=14(k-1)^2.
14(k−1)2=(214)2=56⇒(k−1)2=4⇒k=3 or −1.14(k-1)^2=(2\sqrt{14})^2=56\Rightarrow(k-1)^2=4\Rightarrow k=3\ \text{or}\ -1.
 Step 5: α>0⇒k=3⇒S=(3,7,11)\alpha>0\Rightarrow k=3\Rightarrow S=(3,7,11).
∴α+β+γ=3+7+11=21.\therefore\alpha+\beta+\gamma=3+7+11=21.
 Correct answer: (3)
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