A line passes through $(2,2)$ and is perpendicular to the line $3 x+y=3$ Slope of line $3 \mathrm{x}+\mathrm{y}=3$ is $-3$
Slope of line which passes through $(2,2)$ is $\frac{1}{3}$
$\therefore$ Equation of line passes through $(2,2)$ and having
slope $\left(\frac{1}{3}\right)$ is
$(\mathrm{y}-2)=\frac{1}{3}(\mathrm{x}-2)$
$\therefore \quad 3 y-6=x-2$
$\therefore \quad x-3 y+4=0$
In order to find y-intercept of the line Put $x=0$ in $x-3 y+4=0$ $\therefore \quad-3 \mathrm{y}=-4$
$\mathrm{y}=\frac{4}{3}$
$\therefore$ Option (b) is correct.