JEE Main 2026 — Three Dimensional Geometry Question with Solution

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

$\frac{1-4}{1}=\frac{\mu -9}{2}=\frac{2-5}{1}$

$-3=\frac{\mu -9}{2}=-3\Rightarrow \mu -9=-6\Rightarrow \mu =3$

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

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

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

$1-\lambda +2-1=0\Rightarrow \lambda =2$

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

$L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$

$L_2:\frac{x-2}{2}=\frac{y-3}{3}=\frac{z+5}{6}$

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

The distance $d$ between two parallel lines is given by $d=\frac{|({\vec{a}}_2-{\vec{a}}_1)\times \vec{b}|}{|\vec{b}|}$.

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

Now, we compute the cross product:

$({\vec{a}}_2-{\vec{a}}_1)\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -1\\ 2& 3& 6\end{array}\right|$

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

The magnitude of this vector is:

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

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

$|\vec{b}|=\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$

Therefore, the distance between the lines is:

$d=\frac{\sqrt{146}}{7}$

Answer: $\frac{\sqrt{146}}{7}$