[tex]z=x^2+y^2[/tex]
[tex]x+y+2z=30\implies z=\dfrac{30-x-y}2[/tex]
[tex]\implies x^2+y^2=\dfrac{30-x-y}2\implies x^2+\dfrac x2+y^2+\dfrac y2=15[/tex]
Complete the square to get the equation of the ellipse,
[tex]x^2+\dfrac x2+\dfrac1{16}+y^2+\dfrac y2+\dfrac1{16}=15[/tex]
[tex]\implies\left(x+\dfrac14\right)^2+\left(y+\dfrac14\right)^2=\dfrac{121}8[/tex]
Parameterize this intersection by the vector-valued function,
[tex]\vec r(t)=\langle x(t),y(t),z(t)\rangle[/tex]
[tex]\vec r(t)=\left\langle\sqrt{\dfrac{121}8}\cos t-\dfrac14,\sqrt{\dfrac{121}8}\sin t-\dfrac14,\dfrac{30-\left(\sqrt{\frac{121}8}\cos t-\frac14\right)-\left(\sqrt{\frac{121}8}\sin t-\frac14\right)}2\right\rangle[/tex]
[tex]\vec r(t)=\left\langle\sqrt{\dfrac{121}8}\cos t-\dfrac14,\sqrt{\dfrac{121}8}\sin t-\dfrac14,\dfrac{241}{16}-\sqrt{\dfrac{121}{32}}(\cos t+\sin t)\right\rangle[/tex]
where [tex]0\le t\le2\pi[/tex].
The distance between the origin and any point on the curve is [tex]\|\vec r(t)\|[/tex], which is
[tex]\|\vec r(t)\|=\sqrt{x(t)^2+y(t)^2+z(t)^2}[/tex]
Recall that a function [tex]f(x)[/tex] and its square root [tex]\sqrt{f(x)}[/tex] share the same critical points, so we can make things easier by looking for the critical points of [tex]\|\vec r(t)\|^2[/tex]. Let
[tex]f(t)=\|\vec r(t)\|^2[/tex]
[tex]f(t)=\left(\sqrt{\dfrac{121}8}\cos t-\dfrac14\right)^2+\left(\sqrt{\dfrac{121}8}\sin t-\dfrac14\right)^2+\left(\dfrac{241}{16}-\sqrt{\dfrac{121}{32}}(\cos t+\sin t)\right)^2[/tex]
[tex]f(t)=\dfrac{8051}{32}-\dfrac{693}{8\sqrt2}(\cos t+\sin t)+\dfrac{121}{16}\sin t\cos t[/tex]
Differentiate and find the critical points:
[tex]f'(t)=\dfrac{693}{8\sqrt2}(\sin t-\cos t)+\dfrac{121}{16}(\cos^2t-\sin^2t)[/tex]
[tex]f'(t)=\dfrac1{16\sqrt2}(\sin t-\cos t)(1386+121\sqrt2(\cos t+\sin t))=0[/tex]
[tex]\implies\sin t=\cos t\implies t=\dfrac\pi4,\dfrac{5\pi}4[/tex]
The distance from the origin to the critical points of [tex]\vec r(t)[/tex] are then
[tex]\left\|\vec r\left(\dfrac\pi4\right)\right\|=\dfrac{15\sqrt3}2[/tex]
[tex]\left\|\vec r\left(\dfrac{5\pi}4\right)\right\|=3\sqrt{38}[/tex]
Then the point closest to the origin is
[tex]\left(x\left(\dfrac\pi4\right),y\left(\dfrac\pi4\right),z\left(\dfrac\pi4\right)\right)=\left(\dfrac52,\dfrac52,\dfrac{25}2\right)[/tex]
and the point furthest from the origin is
[tex]\left(x\left(\dfrac{5\pi}4\right),y\left(\dfrac{5\pi}4\right),z\left(\dfrac{5\pi}4\right)\right)=(-3,-3,18)[/tex]
