A line having direction ratios \(1,-4,2\) intersects the lines \(\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}\) and \(\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}\) at the points \(A\) and \(B\) resp., then co-ordinates of points A and B are

Let \(\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}=\lambda\)
\(\Rightarrow x=3 \lambda+7, y=-\lambda+1, z=\lambda-2\)
Let \(\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}=\mu\)
\(\Rightarrow x=2 \mu, y=3 \mu+7, z=\mu\)
Co-ordinates of a point on the first line are
\(\mathrm{A}(3 \lambda+7,1-\lambda, \lambda-2)\)
Co-ordinates of a point on the second line are
\(\mathrm{B}(2 \mu, 3 \mu+7, \mu)\)
D.r.s. of \(A B\) are
\(3 \lambda-2 \mu+7,-\lambda-3 \mu-6, \lambda-\mu-2\)
D.r.s. of \(A B\) are \(1,-4,2\)
\(\begin{aligned}
\therefore \quad & \frac{3 \lambda-2 \mu+7}{1}=\frac{-\lambda-3 \mu-6}{-4}=\frac{\lambda-\mu-2}{2} \\
& 3 \lambda-2 \mu+7=\frac{\lambda+3 \mu+6}{4} \\
& \Rightarrow \lambda-\mu+2=0 ...(i)\\
& \frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2} \\
& \Rightarrow \lambda-5 \mu-10=0...(ii)
\end{aligned}\)
Solving (i) and (ii), we get
\(\begin{aligned}
\lambda & =-5, \mu=-3 \\
\therefore \quad & A \equiv(-8,6,-7), \\
& B \equiv(-6,-2,-3)
\end{aligned}\)