$ \displaystyle \left(i\right)\bar{F}=(x+2y+az)\ i+\left(bx-3y-z\right)j+\left(4x+cy+2z\right)k\ \\[2ex] \therefore{}r=xi+yj+zk\\[2ex] \displaystyle Since,\bar{F}\ is\ irrotational,\ \nabla{}\times{}F=0.\ \\[2ex] \displaystyle Curl\ F=\nabla{}\times{}F=\left\vert{}\begin{array}{ ccc} i & j & k \\ \frac{\partial{}}{\partial{}x} & \frac{\partial{}}{\partial{}y} & \frac{\partial{}}{\partial{}z} \\ \left(x+2y+az\right) & \left(bx-3y-z\right) & \left(4x+cy+2z\right) \end{array}\right\vert{}=0 \\[2ex] $

$ \displaystyle i\left(c-\left(-1\right)\right)-j\left(4-a\right)+k\left(b-2\right)=0i+0j+0k \\[2ex] \displaystyle \therefore{},\ c+1=0,\ 4-a=0\ ,\ b-2=0 \\[2ex] \displaystyle \therefore{},\ a=4,\ b=2,\ c=\ -1 \\[2ex] $