3D Geometry
Equal Distances from Line — a+b+c
nta_pyq_2026_jan
Grade 12
Question:
If the distances of the point $(1,2,a)$ from the line $\dfrac{x-1}{1}=\dfrac{y}{2}=\dfrac{z-1}{1}$ along the lines $L_1:\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-a}{b}$ and $L_2:\dfrac{x-1}{1}=\dfrac{y-2}{4}=\dfrac{z-a}{c}$ are equal, then $a+b+c$ is equal to
Step-by-Step Solution
Key Concept: Point $A=(3\lambda+1,4\lambda+2,b\lambda+a)$ on $L_1$ lies on given line $L$: $3\lambda/1=(4\lambda+1)/2=(b\lambda-1)/1$. $6\lambda=4\lambda+1\Rightarrow\lambda=1/2$... From solution: $a=3$, $b=1$, $c=3$.
$a+b+c=7$.
Correct Answer: 2