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If the line passing through the points $(a, 1,6)$ and $(3,4, b)$ crosses the yz-plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$, then
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The correct answer is:
$a=5, b=1$
Let $\mathrm{y}-\mathrm{z}$ plane divides the join in the ratio $\lambda: 1$
$\begin{aligned} & \Rightarrow\left(\frac{3 \lambda+a}{\lambda+1}, \frac{4 \lambda+1}{\lambda+1}, \frac{b \lambda+6}{\lambda+1}\right) \equiv\left(0 ; \frac{17}{2}, \frac{-13}{2}\right) \\ & \Rightarrow 3 \lambda+a=0, \frac{4 \lambda+1}{\lambda+1}=\frac{17}{2}, \frac{b \lambda+6}{\lambda+1}=\frac{-13}{2} \\ & \Rightarrow \lambda=\frac{-5}{3}, a=5, b=1\end{aligned}$
$\begin{aligned} & \Rightarrow\left(\frac{3 \lambda+a}{\lambda+1}, \frac{4 \lambda+1}{\lambda+1}, \frac{b \lambda+6}{\lambda+1}\right) \equiv\left(0 ; \frac{17}{2}, \frac{-13}{2}\right) \\ & \Rightarrow 3 \lambda+a=0, \frac{4 \lambda+1}{\lambda+1}=\frac{17}{2}, \frac{b \lambda+6}{\lambda+1}=\frac{-13}{2} \\ & \Rightarrow \lambda=\frac{-5}{3}, a=5, b=1\end{aligned}$
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