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MathematicsInteger+4 / -1Question 23 of 75

JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 23

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\mathbf R.
If (a,b,c)(a,b,c), a∈Za\in\mathbf 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 ________.

Answer and worked solution

Official answer: 4

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

Meeting L2L_2. (3+t,2t−1,2t+4)=k(−2,3,−2)(3+t,2t-1,2t+4)=k(-2,3,-2) gives t=−1t=-1, k=−1k=-1: the point (2,−3,2)(2,-3,2).

Point on L3L_3. (3+2s,3+2s,2+s)(3+2s,3+2s,2+s) at squared distance 17:
(1+2s)2+(6+2s)2+s2=17 ⇒ 9s2+28s+20=0,(1+2s)^2+(6+2s)^2+s^2=17\ \Rightarrow\ 9s^2+28s+20=0,
so s=−2s=-2 or −109-\tfrac{10}9. An integer a needs s=−2s=-2: (−1,−1,0)(-1,-1,0).

Answer. (a+b+c)2=4(a+b+c)^2=4.

Topics:Equation of a line in spaceCross product
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