Question Id : 17970 |
Context : JEE Main 2026 (22 January Evening Shift)
Let $L$ be the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z+3}{6}$ and let $S$ be the set of all points $(a,b,c)$ on $L$, whose distance from the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-9}{0}$ along the line $L$ is $7$. Then $\sum_{(a,b,c)\in S}(a+b+c)$ is equal to:
🎥 Video solution / Text Solution of this question is given below:
Parametrize line $L$: $x=-1+2t,y=-1+3t,z=-3+6t$. Distance condition gives two values of $t$. Sum $(a+b+c)$ for both points gives $28$