Given line is $x+y+1=0$.....(i)
Let the foot of $\perp$ of the point $(2,3)$ on (i) be $D(x_1,y_1)$.
Slope of the line $x+y+1=0$ is $m_1=-1$
$\Rightarrow\: $ Slope of the $\perp$ is $1$
$i.e.,m_2=1$ (slope of perpendicular)
$\therefore$ Equation of the perpendicular line passing through $(2,3)$ will be
Now finding the intersection of the lines $y-x=1$ & $x+y+1=0$
Putting (1) & (2)
Hence foot of perpendicular will be $(-1,0)$.
Hence (c) is the correct answer.