Recall that the normal matrix Mn performs model$-$view transformations for surface normals $($by $1$ transforming them from object spaces to camera spaces$).$ Given a model$-$view matrix Mmv $=10000100002100001,$ and a triangle with vertices p$0=(1,0,0)>,$ p$1=(0,1,0)>,$ and p$2=(0,0,1)>$ in the object space, please: $($i$)$ Write down the normal matrix Mn corresponding to Mmv$.($ii$)$ Calculate the object$-$space normal n$$ of the triangle via n$=($p$1$ p$0)($p$2$ p$0)$ k$($p$1$ p$0)($p$2$ p$0)$k $.(1)($Feel free to use a calculator$/$program to compute the cross products.$)($iii$)$ Calculate the camera$-$space normal n$0$ of the triangle based on camera$-$space positions of its vertices. That is$,$ n$0=($p $01$ p $00)($p $02$ p $00)$ k$($p $01$ p $00)($p $02$ p $00)$k $,(2)$ where p $0$ i $=$ Mmv pi for i $=0,1,2($under homogeneous coordinates$).($iv$)$ Calculate the camera$-$space normal n$0$ using the normal matrix Mn and the triangle$$s object$-$space normal n$$ from Eq$.(1)$ via: n$0=$ Mn n$$ kMn n$$k $.$ If your calculations are correct, this result should match the one given by Eq$.($