Is the solution correct? Please check it and tell me my mistakes. Thank you.
Find an equation for each plane tangent to $K$ which is parallel to the plane $x-y+z=5$, where $K$ is the half-cone $z=\sqrt{x^2+y^2}$
Let $f(x,y,z) = z-\sqrt{x^2+y^2}$.
To get the normal vector to the cone at the $(x,y,z)$, calculate the gradient of $f$:
\begin{align*}
\nabla f(x,y,z) &= \frac {2x}{-2\sqrt{x^2 + y^2}}\mathbf i+\frac{2y}{-2\sqrt{x^2+y^2}}\mathbf j+\mathbf k \\
&=-\frac x {\sqrt{x^2+y^2}}\mathbf i – \frac y {\sqrt{x^2+y^2}}\mathbf j + \mathbf k.
\end{align*}
Since we want the tangent plane to be parallel to the plane $x-y+z=5$, the normal vector has to be parallel to the vector $\mathbf i-\mathbf j+\mathbf k$ since this is normal to the plane ($x-y+z=5$). That is,
\begin{gather}
\require{cancel}
-\frac x {\sqrt{x^2+y^2}}\mathbf i
– \frac y {\sqrt{x^2+y^2}}\mathbf j
+ \cancel{\mathbf k} = c(\mathbf i – \mathbf j + \cancel{\mathbf k})
\qquad \text{constant $c = 1$} \\
\left.\begin{gathered}
-\frac x z = 1 \implies \bbox[2.5pt,border:1pt solid black]{x = -z}\\
-\frac y z = 1 \implies \bbox[2.5pt,border:1pt solid black]{y = z}
\end{gathered}\right\}
\:
\begin{gathered}
\land z = \sqrt{x^2+y^2} \implies z = \sqrt{2z^2} \implies z=0 \\
\bbox[2.5pt,border:1pt solid black]{x=0,\,y=0}
\end{gathered}
\end{gather}
So $K$ has no tangent plane at the origin, which is parallel to $x-y+z=5$.
Best Answer
This solution looks correct. However, there is one careless error towards the very end. Where you have: $$-\frac{y}{z} = 1$$ It should read: $$-\frac{y}{z} = -1$$
But, it appears to be a careless error, because you ended up correcting for it.