The correct option is (a) -184
Best explanation: Given that u=x+3y^2-z^3, v=4x^2 yz, w=2z^2-xy
\(\frac{∂(u,v,w)}{∂(x,y,z)} = \begin{vmatrix}
\frac{∂u}{∂x} & \frac{∂u}{∂y} &\frac{∂u}{∂z}\\
\frac{∂v}{∂x} &\frac{∂v}{∂y} &\frac{∂v}{∂z}\\
\frac{∂w}{∂x} &\frac{∂w}{∂y} &\frac{∂w}{∂z}\\
\end{vmatrix} = \begin{vmatrix}
1 & 6y & -3z^2\\
8xyz &4x^2 z &4x^2 y\\
-y &-x &4z\\
\end{vmatrix}\)
\(\frac{∂(u,v,w)}{∂(x,y,z)} \,at\, (1,1,1) = \begin{vmatrix}
1 & 6 & -3\\
8 &4&4\\
-1 &-1 &4\\
\end{vmatrix}\)
=1(16+4) – 6(32+4) – 3(-8+4) = -184.