If a circle, whose centre is $(-1, 1)$ touches the straight line $x + 2y + 12 = 0$, then the coordinates of the point of contact are

The point of contact is the foot of the perpendicular from the centre $(-1,1)$ to the line $x+2y+12=0$. For the line $ax+by+c=0$, the foot of the perpendicular from $(x_1,y_1)$ is $$\left(x_1-\frac{a(ax_1+by_1+c)}{a^2+b^2},\ y_1-\frac{b(ax_1+by_1+c)}{a^2+b^2}\right).$$ Here $a=1$, $b=2$, $c=12$, $(x_1,y_1)=(-1,1)$. $$ax_1+by_1+c=1\cdot(-1)+2\cdot1+12=13$$ $$a^2+b^2=1^2+2^2=5$$ So the foot of the perpendicular is $$\left(-1-\frac{13}{5},\ 1-\frac{26}{5}\right)$$ $$\left(\frac{-18}{5},\frac{-21}{5}\right).$$ Now checking the options, this matches option $\mathrm{B}$.