JEE Main 2026 — Three Dimensional Geometry Question with Solution

The direction vectors of the given lines $L_2$ and $L_3$ are $\overrightarrow{d_2}=\hat{i}+2\hat{j}+2\hat{k}$ and $\overrightarrow{d_3}=2\hat{i}+2\hat{j}+\hat{k}$ respectively.

Since line $L_1$ is perpendicular to both $L_2$ and $L_3$, its direction vector $\overrightarrow{d_1}$ is given by the cross product of $\overrightarrow{d_2}$ and $\overrightarrow{d_3}$:
$\overrightarrow{d_1}=\overrightarrow{d_2}\times \overrightarrow{d_3}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 2& 2& 1\end{array}\right|=-2\hat{i}+3\hat{j}-2\hat{k}$

Since $L_1$ passes through the origin, its equation is:
$\vec{r}=\lambda (-2\hat{i}+3\hat{j}-2\hat{k})$

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

From the first and third equations, we get:
$3+t=2t+4\Rightarrow t=-1$

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

Let the point on $L_3$ be $Q(a,b,c)=(3+2s,3+2s,2+s)$. The distance between $P$ and $Q$ is given as $\sqrt{17}$.
$P{Q}^2=17$
$(3+2s-2{)}^2+(3+2s-(-3){)}^2+(2+s-2{)}^2=17$
$(2s+1{)}^2+(2s+6{)}^2+s^2=17$
$4s^2+4s+1+4s^2+24s+36+s^2=17$
$9s^2+28s+37=17$
$9s^2+28s+20=0$
$9s^2+18s+10s+20=0$
$(9s+10)(s+2)=0$

This gives $s=-2$ or $s=-\frac{10}{9}$.

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

Substituting $s=-2$ into the coordinates of $Q$:
$a=3+2(-2)=-1$
$b=3+2(-2)=-1$
$c=2+(-2)=0$

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

We need to find the value of $(a+b+c{)}^2$:
$(a+b+c{)}^2=(-1-1+0{)}^2=(-2{)}^2=4$

Answer: $4$