JEE Challenger
More from Three Dimensional Geometry
MathMedMain 2026 • 8th Apr - Shift 2

Distance of Point on Line L3 From Intersection of L1 and L2

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∈Z(a, b, c), a \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 ________.

Official Numerical Answer4

Topics & Concepts

Three Dimensional GeometryLine in 3D

Step-by-Step Solution

To find the required value, we determine the direction vector of L1L_1 by taking the cross product of the direction vectors of L2L_2 and L3L_3: d⃗1=(i^+2j^+2k^)×(2i^+2j^+k^)=−2i^+3j^−2k^\vec{d}_1 = (\hat{i} + 2\hat{j} + 2\hat{k}) \times (2\hat{i} + 2\hat{j} + \hat{k}) = -2\hat{i} + 3\hat{j} - 2\hat{k}

Since L1L_1 passes through the origin, its parametric form is r⃗=−2ui^+3uj^−2uk^\vec{r} = -2u\hat{i} + 3u\hat{j} - 2u\hat{k}. Equating this with L2L_2, we find the point of intersection of L1L_1 and L2L_2 to be P(2,−3,2)P(2, -3, 2).

A general point on L3L_3 is given by (3+2s,3+2s,2+s)(3+2s, 3+2s, 2+s). Using the distance formula between this point and P(2,−3,2)P(2, -3, 2): ((3+2s)−2)2+((3+2s)+3)2+((2+s)−2)2=17((3+2s) - 2)^2 + ((3+2s) + 3)^2 + ((2+s) - 2)^2 = 17 9s2+28s+20=0  ⟹  (s+2)(9s+10)=09s^2 + 28s + 20 = 0 \implies (s + 2)(9s + 10) = 0

Given a=3+2s∈Za = 3 + 2s \in \mathbb{Z}, we select s=−2s = -2, which gives: a=−1,b=−1,c=0a = -1, \quad b = -1, \quad c = 0

Thus, (a+b+c)2=(−1−1+0)2=4(a + b + c)^2 = (-1 - 1 + 0)^2 = 4.

Distance of Point on Line L3 From Intersection of L1 and L2 | Mathematics PYQ Solution - JEE Challenger