Answer
$$K = 0,{\text{ }}r = \frac{1}{K},{\text{ is undefined}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& y = 3x - 2 \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ so}} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {3x - 2} \right] \cr
& y'\left( x \right) = 3 \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ 3 \right] \cr
& y''\left( x \right) = 0 \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& K = \frac{{\left| 0 \right|}}{{{{\left( {1 + {{\left[ 3 \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Simplify}} \cr
& K = 0 \cr
& {\text{Evaluate at }}x = a \cr
& K = 0 \cr
& {\text{The radius of curvature is }}r = \frac{1}{K},{\text{ (undefined)}}{\text{.}} \cr
& \cr
& K = 0,{\text{ }}r = \frac{1}{K},{\text{(undefined)}}{\text{.}} \cr} $$
