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

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

$$0=9x^2+16y^2+4z^2-2(6xy+4yz+3zx)=\frac{1}{2}((3x-4y)^2+(3x-2z)^2+(4y-2z)^2),$$
which gives $$3x=4y=2z.$$
Now, let $3x=t.$ 
Thus, $$t^2+t^2+t^2=108$$ or $$t^2=36,$$ which gives
$$x^2+y^2+z^2=\frac{t^2}{9}+\frac{t^2}{16}+\frac{t^2}{4}=4+\frac{9}{4}+9=\frac{61}{4}.$$
