You essentially want to find the circumcenter $D$ of the triangle $\triangle_{ABC}$ with vertices
$$A = (-1,0,0), B = (0,2,0), C = (0,0,3)$$
Since $D$ is lying on the plane holding $A, B, C$, there exists $3$ real numbers
$\alpha,\beta,\gamma$ such that
$$D = \alpha A + \beta B + \gamma C\quad\text{ with }\quad \alpha + \beta + \gamma = 1$$
This $3$-tuple is called the baricenteric coordinate of $D$. They can be computed using the sides $a,b,c$ of the triangle alone:
$$\alpha : \beta : \gamma \;=\; a^2(-a^2 + b^2 + c^2) : b^2( a^2 - b^2 + c^2) : c^2(a^2 + b^2 - c^2)$$
For the triangle at hand, we have $(a^2,b^2,c^2) = (13,10,5)$. This leads to
$$\alpha : \beta : \gamma \;=\;
13(-13+10+5) : 10(13-10+5) : 5(13+10-5) = 26 : 80 : 90$$
As a result,
$\displaystyle\;D = \frac{26 A + 80 B + 90C}{26 + 80 + 90} = \left(-\frac{35}{98},\frac{40}{49},\frac{135}{98} \right).$
HINT:
The idea is: any solution of the system
$$f=0\\
g=0$$
is also a solution of $f+ \lambda g =0$ for any $\lambda$. So now consider the sphere
of equation
$$x^2 + y^2 +z^2 + 2 a x + 2 b y + 2 c z + d + \lambda(A x + B y + C z + D)=0$$
that passes through the point $(x_0, y_0, z_0)$. $\lambda$ can be easily obtained now.
Best Answer
This is called a proof by contradiction.
You first do the assumption that your sphere intersects the plane. From this you would deduce that $(x-2)^2 + (z-4)^2 = -11$. But this is not possible since the number on the left is non-negative and the one on the right is negative. Hence you deduce that your initial assumption was false: the sphere does not intersect the plane.