A unit normal vector to the surface $$2x^2y+3yz=4$$ at $$(1,-1,-2)$$ is​

Correct option is A

​$$\textbf{Given surface: } F(x, y, z) = 2x^2 y + 3yz \\[6pt]\textbf{Gradient (normal vector): } \nabla F = \left( \frac{\partial F}{\partial x} \right) \hat{i} + \left( \frac{\partial F}{\partial y} \right) \hat{j} + \left( \frac{\partial F}{\partial z} \right) \hat{k} \\[8pt]\frac{\partial F}{\partial x} = 4xy, \quad \frac{\partial F}{\partial y} = 2x^2 + 3z, \quad \frac{\partial F}{\partial z} = 3y \\[8pt]\text{At } (1, -1, -2): \\\frac{\partial F}{\partial x} = -4, \quad \frac{\partial F}{\partial y} = -4, \quad \frac{\partial F}{\partial z} = -3 \\[6pt]\nabla F = -4\hat{i} -4\hat{j} -3\hat{k} \\[6pt]|\nabla F| = \sqrt{(-4)^2 + (-4)^2 + (-3)^2} = \sqrt{41} \\[6pt]\textbf{Unit normal vector: } \frac{1}{\sqrt{41}}(-4\hat{i} -4\hat{j} -3\hat{k})$$

However,unit normal vectors are often expressed with a positive leading coefficient(by convention). Multiplying by −1

​$$\frac{1}{\sqrt{41}} \left( +4\hat{i} + 4\hat{j} + 3\hat{k} \right)$$​​

n=141(4i^+4j^+3k^)

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