JEE Main 2026 — Mathematics Vectors & 3D Geometry

Let the coordinates of $P$P on the first line be $(2\lambda, 3\lambda, 3\lambda - 1)$(2λ,3λ,3λ−1).

Let the coordinates of $Q$Q on the second line be $(-\mu - 1, \mu + 2, 4\mu)$(−μ−1,μ+2,4μ).

The direction ratios of the line segment $PQ$PQ are proportional to $1, -1, 2$1,−1,2. Thus, we can write:

$$\dfrac{-\mu - 1 - 2\lambda}{1} = \dfrac{\mu + 2 - 3\lambda}{-1} = \dfrac{4\mu - 3\lambda + 1}{2} = k$$1−μ−1−2λ​=−1μ+2−3λ​=24μ−3λ+1​=k

This gives the system of equations:

$$-\mu - 2\lambda - 1 = k$$−μ−2λ−1=k

$$\mu - 3\lambda + 2 = -k$$μ−3λ+2=−k

$$4\mu - 3\lambda + 1 = 2k$$4μ−3λ+1=2k

Adding the first two equations gives:

$$-5\lambda + 1 = 0 \Rightarrow \lambda = \dfrac{1}{5}$$−5λ+1=0⇒λ=51​

Substituting $\lambda = \dfrac{1}{5}$λ=51​ into the first and third equations:

$$-\mu - \dfrac{7}{5} = k \Rightarrow \mu + k = -\dfrac{7}{5}$$−μ−57​=k⇒μ+k=−57​

$$4\mu + \dfrac{2}{5} = 2k \Rightarrow 2\mu - k = -\dfrac{1}{5}$$4μ+52​=2k⇒2μ−k=−51​

Adding these two equations gives:

$$3\mu = -\dfrac{8}{5} \Rightarrow \mu = -\dfrac{8}{15}$$3μ=−58​⇒μ=−158​

Substituting $\mu = -\dfrac{8}{15}$μ=−158​ into $\mu + k = -\dfrac{7}{5}$μ+k=−57​:

$$-\dfrac{8}{15} + k = -\dfrac{21}{15} \Rightarrow k = -\dfrac{13}{15}$$−158​+k=−1521​⇒k=−1513​

The vector $\overrightarrow{PQ}$PQ​ is $(k, -k, 2k)$(k,−k,2k). The length of the line segment $PQ$PQ is $\alpha = \sqrt{k^2 + (-k)^2 + (2k)^2} = \sqrt{6k^2}$α=k2+(−k)2+(2k)2​=6k2​.

Thus, $\alpha^2 = 6k^2 = 6 \left(-\dfrac{13}{15}\right)^2 = 6 \times \dfrac{169}{225} = \dfrac{1014}{225}$α2=6k2=6(−1513​)2=6×225169​=2251014​.

Therefore, $225\alpha^2 = 1014$225α2=1014.

Answer: $1014$1014