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At what point on the surface $y = x^2 + z^2$ is the tangent plane parallel to the plane $x + 2y + 3z = 1$?

(Note: two planes are parallel if they have parallel normal vectors)

Thank you so much.

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    I already know the normal vector for $x+2y+3z=1$ is $(1,2,3)$, and the equation for tangent plane is $\langle \nabla F(\mathbf x_0), \mathbf x - \mathbf x_0\rangle=0$. – 2017-02-12

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Getting a normal to the given plane is the easy part. It's given by $(1,2,3)$. Now we want to know when the tangent plane has the same normal.

Now it's a fact that you will probably want to know that if a surface is defined by $f(x,y,z) = 0$, then $\nabla f$ will allow us to find a normal. In this case, if we rewrite your surface as $y - x^2 - z^2 = 0$, then we can look at the gradient. This is easily computed to be $(-2x, 1, -2z)$. Since we're only interested in the direction of the normal and don't care about it's magnitude, let's scale the normals so they have the same $y$ components (this amounts to multiplying both sides of the equation $f(x,y,z) = 0$ by $2$). Now considering the new normal $(-4x, 2, -4z)$, we just have to figure out what $x$ and $z$ should be to make them the same as the normal to the plane. We can easily see $x = -1/4$ and $z = -3/4$ do the job.

This corresponds with the value of $y = 1/16 + 9/16 = 5/8$. So the surface point with the desired property is $(-1/4, 5/8, -3/4)$.

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    Why do we need to know if the surface is defined by $f(x,y,z)$? I thought $\nabla f(x,y,z)$ is always perpendicular to tangent plane at $(x,y,z)$. – 2017-02-12
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    Well the tangent plane of what? The idea is that the equation $f(x,y,z) = k$ represents a surface. If you wrote the surface down some other way, you could still use the gradient to recover the normal, but it won't be in the exact same form - it may be a gradient of just two variables, for example. See if this link doesn't help clear it up. http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx – 2017-02-12