If point $P(a,b,c)$ satisfies $a-b+c=0$ and $a^2+b^2+c^2=4$. The point $P$ traces a plane figure whose area is $A$. Then $\lfloor A\rfloor $ is
solution I try
Let an equation of plane is $ax+by+cz+d=0$. Now distance of point $P(a,b,c)$ from origin is $$\frac{d}{\sqrt{a^2+b^2+c^2}}=\frac{d}{2}$$
How do I solve that problem? Help me.
In, the $a$-$b$-$c$ 3-dimensional Euclidean space (not $x$-$y$-$z$), the second condition $a^2+b^2+c^2=4$ is (the surface of) a sphere with radius $2$.
The first condition $a - b + c = 0$ is a plane that passes through the origin $(a,b,c) = (0,0,0)$, which is also the center of the sphere.
Therefore, the intersection (where both conditions are satisfied) is a great circle of the said sphere, hence its radius is also $r = 2$.
The area of such circle is $A = \pi r^2 = 4\pi \approx 12.56637\ldots$, and the floor of the area $\lfloor A\rfloor$ (I'm not sure why you want that) is thus $12$.
