A line drawn from the point \(\mathrm{A}(1,3,2)\) parallel to the line \(\frac{x}{2}=\frac{y}{4}=\frac{z}{1}\), intersects the plane \(3 x+y+2 z=5\) in point \(\mathrm{B}\), then co-ordinates of point \(\mathrm{B}\) are

The d.r.s. of the line \(\frac{x}{2}=\frac{y}{4}=\frac{z}{1}\) are \(2,4,1\).
\(\therefore \quad\) The d.r.s. of any line parallel to it are also \(2,4,1\). The equation of the line passing through \(\mathrm{A}(1,3,2)\) is \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{1}=\lambda\) (say) Then, any point on the line is \(\mathrm{B}=(2 \lambda+1,4 \lambda+3, \lambda+2)\)
The point \(\mathrm{B}\) lies on the plane \(3 x+y+2 z=5\).
\(\therefore 3(2 \lambda+1)+4 \lambda+3+2(\lambda+2)=5\)
\(\Rightarrow 12 \lambda+10=5\)
\(\Rightarrow \lambda=\frac{-5}{12} \\
\therefore B=\left(\frac{1}{6}, \frac{4}{3}, \frac{19}{12}\right)\)