JEE Main 2026 — Mathematics Vectors & 3D Geometry

Since the point $(1, \mu, 2)$(1,μ,2) lies on the line $\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}$1x−4​=2y−9​=1z−5​, we substitute its coordinates into the line equation:

$$\dfrac{1-4}{1} = \dfrac{\mu-9}{2} = \dfrac{2-5}{1}$$11−4​=2μ−9​=12−5​

$$-3 = \dfrac{\mu-9}{2} = -3 \Rightarrow \mu - 9 = -6 \Rightarrow \mu = 3$$−3=2μ−9​=−3⇒μ−9=−6⇒μ=3

The direction ratios of the given line are $(1, 2, 1)$(1,2,1). The direction ratios of the perpendicular from $(\lambda, 2, 3)$(λ,2,3) to $(1, 3, 2)$(1,3,2) are $(1-\lambda, 3-2, 2-3) = (1-\lambda, 1, -1)$(1−λ,3−2,2−3)=(1−λ,1,−1).

Since the lines are perpendicular, their dot product is zero:

$$1(1-\lambda) + 2(1) + 1(-1) = 0$$1(1−λ)+2(1)+1(−1)=0

$$1 - \lambda + 2 - 1 = 0 \Rightarrow \lambda = 2$$1−λ+2−1=0⇒λ=2

Substituting $\lambda = 2$λ=2 and $\mu = 3$μ=3 into the equations of the two lines, we get:

$$L_1: \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}$$L1​:2x−1​=3y−2​=6z+4​

$$L_2: \dfrac{x-2}{2} = \dfrac{y-3}{3} = \dfrac{z+5}{6}$$L2​:2x−2​=3y−3​=6z+5​

These are parallel lines with direction vector $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$b=2i^+3j^​+6k^. The lines pass through the points $\vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}$a1​=i^+2j^​−4k^ and $\vec{a}_2 = 2\hat{i} + 3\hat{j} - 5\hat{k}$a2​=2i^+3j^​−5k^ respectively.

The distance $d$d between two parallel lines is given by $d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$d=∣b∣∣(a2​−a1​)×b∣​.

We have $\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = \hat{i} + \hat{j} - \hat{k}$a2​−a1​=(2−1)i^+(3−2)j^​+(−5−(−4))k^=i^+j^​−k^.

Now, we compute the cross product:

$$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix}$$(a2​−a1​)×b=​i^12​j^​13​k^−16​​

$$= \hat{i}(6 - (-3)) - \hat{j}(6 - (-2)) + \hat{k}(3 - 2) = 9\hat{i} - 8\hat{j} + \hat{k}$$=i^(6−(−3))−j^​(6−(−2))+k^(3−2)=9i^−8j^​+k^

The magnitude of this vector is:

$$|9\hat{i} - 8\hat{j} + \hat{k}| = \sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81 + 64 + 1} = \sqrt{146}$$∣9i^−8j^​+k^∣=92+(−8)2+12​=81+64+1​=146​

The magnitude of the direction vector $\vec{b}$b is:

$$|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$∣b∣=22+32+62​=4+9+36​=49​=7

Therefore, the distance between the lines is:

$$d = \dfrac{\sqrt{146}}{7}$$d=7146​​

Answer: $\dfrac{\sqrt{146}}{7}$7146​​