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If the line segment joining the points $P(2,4,1)$ and $Q(3,8,1)$ is divided by the plane $3 x-k y-6 z=0$ externally in the ratio $4: 5$, then $k=$
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Verified Answer
The correct answer is:
1
Any point which divide the Ine segment joining the points $P(2,4,1)$ and $Q(3,8,1)$ extemally $4: 5$ is given by
$$
\begin{aligned}
& \left(\frac{4 \times 3-5 \times 2}{4-5}, \frac{4 \times 8-5 \times 4}{4-5}, \frac{4 \times 1-5 \times 1}{4-5}\right) \text { i.e. } \\
& (-2,12,1)
\end{aligned}
$$
Since $(-2,-121)$ lines on the plane $3 x-k y-6 z=0$
$$
\begin{aligned}
& \therefore \quad 3(-2)-k(-12)-6(1)=0 \\
& \Rightarrow-6+12 k-6=0 \Rightarrow 12 k=12 \Rightarrow k=1
\end{aligned}
$$
$$
\begin{aligned}
& \left(\frac{4 \times 3-5 \times 2}{4-5}, \frac{4 \times 8-5 \times 4}{4-5}, \frac{4 \times 1-5 \times 1}{4-5}\right) \text { i.e. } \\
& (-2,12,1)
\end{aligned}
$$
Since $(-2,-121)$ lines on the plane $3 x-k y-6 z=0$
$$
\begin{aligned}
& \therefore \quad 3(-2)-k(-12)-6(1)=0 \\
& \Rightarrow-6+12 k-6=0 \Rightarrow 12 k=12 \Rightarrow k=1
\end{aligned}
$$
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