A line drawn from the point $\mathrm{A}(1,3,2)$ parallel to the line $\frac{x}{2}=\frac{y}{4}=\frac{z}{1}$,…
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 $3 x+y+2 z=5$.
$\begin{array}{ll}
\therefore \quad & 3(2 \lambda+1)+4 \lambda+3+2(\lambda+2)=5 \\
& \Rightarrow 12 \lambda+10=5 \\
& \Rightarrow \lambda=\frac{-5}{12} \\
\therefore \quad & B=\left(\frac{1}{6}, \frac{4}{3}, \frac{19}{12}\right)
\end{array}$