The co-ordinates of the point, where the line through \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) crosses the \(\mathrm{XZ}\)-plane, are

Let \(\mathrm{A}\left(x_1, y_1, \mathrm{z}_1\right)=\mathrm{A}(3,4,1)\) and
\(\mathrm{B}\left(x_2, y_2, z_2\right)=\mathrm{B}(5,1,6)\)
The equation of the line passing through the - points \(\left(x_1^1, y_1, z_1\right)\) and \(\left(x_2, y_2, z_2\right)\) is given by
\(\begin{aligned}
& \therefore \quad \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} \\
& \therefore \quad \frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1^1}{6-1} \\
& \therefore \quad \frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}
\end{aligned}\)
Since the line crosses the \(\mathrm{XZ}\) plane, \(y=0\)
\(\begin{aligned}
\therefore \quad & \frac{x-3}{2}=\frac{4}{3}=\frac{z-1}{5} \\
\therefore \quad \frac{x-3}{2} & =\frac{4}{3} \text { and } \frac{z-1}{5}=\frac{4}{3} \\
\Rightarrow x & =\frac{17}{3} \text { and } \mathrm{z}=\frac{23}{3}
\end{aligned}\)
\(\therefore \quad\) The required point is \(\left(\frac{17}{3}, 0, \frac{23}{3}\right)\).