I can figure out how to plot this for calc 3.

Question:

>a) Plot the circle of radius 3 centered at the point $\{-1, 1, 1\}$ in the plane whose xyz-equation is $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Include in your plot a big enough piece of the plane to accommodate the circle.

My answer:

We are given:

* Radius = $3$
* Center = $\{-1, 1, 1\}$

Then:

$2(x + 1) + 3(y - 1) + (z - 1) = 0$ so

$2x + 2 + 3y - 3 + z - 1 = 0$ so

$2x + 3y + z = -2 + 3 + 1$ so

$2x + 3y + z = 2$

Thus: Normal = $\{2, 3, 1\}$

Equation of line:

$r(t) = \{-1, 1, 1\} + t \{2, 3, 1\}$

Equation of plane:

$2x + 3y + z = 2$ so

$\left(x + \frac{3}{2}y + \frac{z}{2}\right) = 1$

So we can find that the:

* x intercept = $1$
* y intercept = $\frac{2}{3}$
* z intercept = $2$

How can I plot this?

> b) Here's a plot of a spiral in the xy-plane: 

>     Clear[spiral, t];
>     spiral[t_] = {t Cos[2 t], t Sin[2 t]};
>     ParametricPlot[spiral[t], {t, 0, 3 Pi}, AxesLabel -> {"x", "y"}]

>Use your answer to part a) above to help plot a true scale duplicate copy of this spiral on the plane with xyz-equation $2(x + 1) + 3(y - 1) + (z - 1) = 0$. Center your spiral at $\{-1, 1, 1\}$ and include in your plot a big enough hunk of the plane to accommodate the spiral.

How do I plot this?