The number of points on the line 3x + 4y =5, which are at a distance of ${\sec }^2\mathrm{\theta }+2{\mathrm{cosec}}^2\mathrm{\theta },\mathrm{\theta }\in \mathrm{R}$  from the point (1, 3) is

The perpendicular distance of (1, 3) from the line 3x + 4y=5 is 2 units while

${\sec }^2\mathrm{\theta }+2{\mathrm{cosec}}^2\mathrm{\theta }\ge 3\left({\text{ as sec }}^2\mathrm{\theta },{\mathrm{cosec}}^2\mathrm{\theta }\ge 1\right)$

So, there will be two such points on the line.