[shahkb4]

Let L be the line $\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}+3}{6}$ and let S be the set of all points (a,b,c) on L, whose distance from the line $\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-9}{0}$ along the line L is 7. Then $\sum_{(\mathrm{a}, \mathrm{b}, \mathrm{c}) \in \mathrm{S}}(\mathrm{a}+\mathrm{b}+\mathrm{c})$ is equal to
[JEE MAIN 22 Jan 2026 S II]
(1) 34
(2) 28
(3) 40
(4) 6

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[shahkb4]

Solution

line $\mathrm{L} \equiv \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}+3}{6}=\lambda=\mathrm{t}$

\[ \begin{aligned} & \mathrm{A}(2 \lambda-1,3 \lambda-1,6 \lambda-3) \\ & \mathrm{L}_{1}: \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-9}{0}=\mu \\ & \mathrm{A}(2 \mu-1,3 \mu-1,9) \\ & 2 \lambda-1=2 \mu-1 \Rightarrow \lambda=\mu \\ & 6 \lambda-3=9 \Rightarrow 6 \lambda=12 \Rightarrow \lambda=2 \\ & \mathrm{~A}(3,5,9) \end{aligned} \]

Point $\mathrm{P}(2 \mathrm{t}-1,3 \mathrm{t}-1,6 \mathrm{t}-3)$

\[ \left.\begin{array}{l} A P=7 \\ \Rightarrow(2 t-4)^{2}+(3 t-6)^{2}+(6 t-12)^{2}=49 \\ \Rightarrow(t-1)(t-3)=0 \Rightarrow t=1,3 \\ t=1 \quad P_{1}(1,2,3) \\ t=3 \quad P_{2}(5,8,15) \end{array}\right] . \sum_{a, b, c \in S}(a+b+c)=1+2+3+5+8+15=34 \text { Answer }(1) \text {. } \]

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