$\displaystyle r_1 (5 \ 1)+\lambda (3 \ -2)$ and $\displaystyle r_2 (-2 \ 2)+t (4 \ 1)$

presume one way to do this is turn these into line equations

$\displaystyle y=-\frac{2}{3}x+6$

$\displaystyle y=\frac{1}{4}x+\frac{3}{2}$

then $\displaystyle x=\frac{46}{3}$ and $\displaystyle y=\frac{16}{3}$