Let L be the line (x+1)/(2)=(y+1)/(3)=(z+3)/(6) and let S be the set of all points (a, b, …

Question

Let L be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ on L , whose distance from the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line $L$ is 7 ...

Accepted Answer

M is the point of intersection of ${L}_{1}\&{L}_{2}$

$\begin{gathered} \Rightarrow 2\lambda -1=2\mu -1,3\lambda -1=3\mu -1,6\lambda -3=9\Rightarrow \lambda =2=\mu \ M(3,5,9) \end{gathered}$

Now let point P be $(2K-1,3K-1,6K-3)$ on ${L}_{2}$

Such that $PM=7$

$\begin{gathered} \begin{gathered} \quad \Rightarrow \sqrt{(2K-4{)}^{2}+(3K-6)+(6K-12{)}^{2}}=7 \ \Rightarrow 49{K}^{2}+196-196K=49 \ \Rightarrow {K}^{2}+4-4K=1\Rightarrow {K}^{2}-4K+3=0\Rightarrow K=1,3 \end{gathered} \ P\quad Q(1,2,3)\quad (5,8,15)P\quad Q=34 \end{gathered}$

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