For real numbers $x,y$ and $z$ if $9x^2+16y^2+4z^2=108$ and $6xy+4yz+3zx=54$. Find value of $x^2+y^2+z^2$.

Question

Question

For real numbers $x,y$ and $z$ if $9x^2+16y^2+4z^2=108$ and $6xy+4yz+3zx=54$. Find value of $x^2+y^2+z^2$.

My calculations

I found $$(3x+4y+2z)^2=108+4×54=324$$ I have no clue how to proceed further .

But for check of validity of question I let wolfram solve it . But results left me more puzzled .

Now how can I get to this result . There is one more picture

Now why I don't get same result $x^2+y^2+z^2$

Feel free to edit , comment and advise . Thank You

Answer 1

Let $a=3x,b=4y,c=2z$. Then we have $a^2+b^2+c^2=108$ and $\frac{ab+bc+ac}2=54$ i.e. $$a^2+b^2+c^2=ab+bc+ac=108,$$ which implies $$a^2+b^2+c^2-ab-bc-ac=\frac{(a-b)^2+(b-c)^2+(c-a)^2}2=0$$ and thus $a=b=c$ and since $a^2+b^2+c^2=108$, we have $a^2=b^2=c^2=\frac{108}3=36$ so that $$x^2+y^2+z^2=a^2\cdot\left(\frac19+\frac1{16}+\frac14\right)=\frac{61}4.$$

Answer 2

$$0=9x^2+16y^2+4z^2-2(6xy+4yz+3zx)=\frac{1}{2}((3x-4y)^2+(3x-2z)^2+(4y-2z)^2),$$ which gives $$3x=4y=2z.$$ Now, let $3x=t.$

Thus, $$t^2+t^2+t^2=108$$ or $$t^2=36,$$ which gives $$x^2+y^2+z^2=\frac{t^2}{9}+\frac{t^2}{16}+\frac{t^2}{4}=4+\frac{9}{4}+9=\frac{61}{4}.$$