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JEE MAIN 2026 · Three Dimensional Geometry

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 23

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

Let a line L1L_1 pass through the origin and be perpendicular to the lines

L2:r⃗=(3+t)i^+(2t−1)j^+(2t+4)k^L_2: \vec r = (3 + t)\hat i + (2t - 1)\hat j + (2t + 4)\hat k and

L3:r⃗=(3+2s)i^+(3+2s)j^+(2+s)k^L_3: \vec r = (3 + 2s)\hat i + (3 + 2s)\hat j + (2 + s)\hat k, t,s∈Rt, s \in \mathbb{R}.

If (a,b,c)(a, b, c), a∈Za \in \mathbb{Z}, is the point on L3L_3 at a distance of 17\sqrt{17} from the point of intersection of L1L_1 and L2L_2, then (a+b+c)2(a + b + c)^2 is equal to ______.

Your answer

Solution

Solution

Direction of L1L_1. (1,2,2)×(2,2,1)=(2−4,  4−1,  2−4)=(−2,3,−2)(1, 2, 2) \times (2, 2, 1) = (2 - 4,\; 4 - 1,\; 2 - 4) = (-2, 3, -2).

Intersection of L1L_1 and L2L_2. Need (3+t,2t−1,2t+4)=k(−2,3,−2)(3 + t, 2t - 1, 2t + 4) = k(-2, 3, -2). The first and third give 3+t=2t+43 + t = 2t + 4, so t=−1t = -1: the point is P(2,−3,2)P(2, -3, 2), and k=−1k = -1 checks the middle coordinate (−3-3).

Point on L3L_3. Q=(3+2s,3+2s,2+s)Q = (3 + 2s, 3 + 2s, 2 + s), so PQ⃗=(1+2s,6+2s,s)\vec{PQ} = (1 + 2s, 6 + 2s, s):

(1+2s)2+(6+2s)2+s2=17  ⇒  9s2+28s+20=0  ⇒  s=−2 or s=−109.(1 + 2s)^2 + (6 + 2s)^2 + s^2 = 17 \;\Rightarrow\; 9s^2 + 28s + 20 = 0 \;\Rightarrow\; s = -2 \text{ or } s = -\tfrac{10}{9}.

s=−109s = -\frac{10}{9} gives a=79a = \frac{7}{9}, not an integer. s=−2s = -2 gives (a,b,c)=(−1,−1,0)(a, b, c) = (-1, -1, 0), and (a+b+c)2=4(a + b + c)^2 = 4.

Answer: 4.

Answer checked against the official answer key. The solution is our own working.

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