JEE Main 2026 — Mathematics Vectors & 3D Geometry

The direction vectors of the given lines $L_2$L2​ and $L_3$L3​ are $\vec{d_2} = \hat{i} + 2\hat{j} + 2\hat{k}$d2​​=i^+2j^​+2k^ and $\vec{d_3} = 2\hat{i} + 2\hat{j} + \hat{k}$d3​​=2i^+2j^​+k^ respectively.

Since line $L_1$L1​ is perpendicular to both $L_2$L2​ and $L_3$L3​, its direction vector $\vec{d_1}$d1​​ is given by the cross product of $\vec{d_2}$d2​​ and $\vec{d_3}$d3​​:

$$\vec{d_1} = \vec{d_2} \times \vec{d_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = -2\hat{i} + 3\hat{j} - 2\hat{k}$$d1​​=d2​​×d3​​=​i^12​j^​22​k^21​​=−2i^+3j^​−2k^

Since $L_1$L1​ passes through the origin, its equation is:

$$\vec{r} = \lambda(-2\hat{i} + 3\hat{j} - 2\hat{k})$$r=λ(−2i^+3j^​−2k^)

Let the point of intersection of $L_1$L1​ and $L_2$L2​ be $P$P. The coordinates of $P$P can be written as $(-2\lambda, 3\lambda, -2\lambda)$(−2λ,3λ,−2λ) from $L_1$L1​ and $(3+t, 2t-1, 2t+4)$(3+t,2t−1,2t+4) from $L_2$L2​. Equating the respective coordinates:

$$-2\lambda = 3 + t$$−2λ=3+t

$$3\lambda = 2t - 1$$3λ=2t−1

$$-2\lambda = 2t + 4$$−2λ=2t+4

From the first and third equations, we get:

$$3 + t = 2t + 4 \Rightarrow t = -1$$3+t=2t+4⇒t=−1

Substituting $t = -1$t=−1 into the first equation gives $-2\lambda = 2 \Rightarrow \lambda = -1$−2λ=2⇒λ=−1. This also satisfies the second equation. Thus, the point of intersection is $P(2, -3, 2)$P(2,−3,2).

Let the point on $L_3$L3​ be $Q(a, b, c) = (3+2s, 3+2s, 2+s)$Q(a,b,c)=(3+2s,3+2s,2+s). The distance between $P$P and $Q$Q is given as $\sqrt{17}$17​.

$$PQ^2 = 17$$PQ2=17

$$(3+2s - 2)^2 + (3+2s - (-3))^2 + (2+s - 2)^2 = 17$$(3+2s−2)2+(3+2s−(−3))2+(2+s−2)2=17

$$(2s + 1)^2 + (2s + 6)^2 + s^2 = 17$$(2s+1)2+(2s+6)2+s2=17

$$4s^2 + 4s + 1 + 4s^2 + 24s + 36 + s^2 = 17$$4s2+4s+1+4s2+24s+36+s2=17

$$9s^2 + 28s + 37 = 17$$9s2+28s+37=17

$$9s^2 + 28s + 20 = 0$$9s2+28s+20=0

$$9s^2 + 18s + 10s + 20 = 0$$9s2+18s+10s+20=0

$$(9s + 10)(s + 2) = 0$$(9s+10)(s+2)=0

This gives $s = -2$s=−2 or $s = -\dfrac{10}{9}$s=−910​.

Since $a = 3 + 2s$a=3+2s must be an integer ($a \in \mathbb{Z}$a∈Z), we must choose $s = -2$s=−2.

Substituting $s = -2$s=−2 into the coordinates of $Q$Q:

$$a = 3 + 2(-2) = -1$$a=3+2(−2)=−1

$$b = 3 + 2(-2) = -1$$b=3+2(−2)=−1

$$c = 2 + (-2) = 0$$c=2+(−2)=0

Therefore, $(a, b, c) = (-1, -1, 0)$(a,b,c)=(−1,−1,0).

We need to find the value of $(a+b+c)^2$(a+b+c)2:

$$(a+b+c)^2 = (-1 - 1 + 0)^2 = (-2)^2 = 4$$(a+b+c)2=(−1−1+0)2=(−2)2=4

Answer: $4$4