Answer
Vector opposite the gradient: $\mathbf{i}-11 \mathbf{j}-12 \mathbf{k}=\mathbf{V}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{266}}(\mathbf{i}-11 \mathbf{j}-12 \mathbf{k})$
Rate of change: $-\sqrt{266}=-\|\nabla f(5,7,6)\|$
Work Step by Step
$f$ increases more quickly in the positive direction of $\nabla f(x, y)$
\[
\begin{array}{l}
\frac{1}{z-y}=f_{x}(x, y, z) \\
-1=f_{x}(5,7,6) \\
\frac{x+z}{(z-y)^{2}}=f_{x}(x, y, z) \\
11=f_{x}(5,7,6) \\
-\frac{x+y}{(z-y)^{2}}=f_{x}(x, y, z) \\
-12=f_{x}(5,7,6)
\end{array}
\]
$-\mathbf{i}+11 \mathbf{j}-12 \mathbf{k}=\nabla f(5,7,6)$
Vector opposite the gradient: $\mathbf{i}-11 \mathbf{j}-12 \mathbf{k}=\mathbf{V}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{266}}(\mathbf{i}-11 \mathbf{j}-12 \mathbf{k})$
Rate of change: $-\sqrt{266}=-\|\nabla f(5,7,6)\|$
