I'm having trouble finding the values of $u$, $v$ and $w$ through solving the equations by substituting the given points in $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.
The answer given is $2(x^2+y^2+z^2)- 3x + 2y -13z + 15 = 0$ and my answer is nowhere close.
Edit: Perhaps it may be either the equation is wrong or the answer is. I believe it more likely to be an incorrect answer.
First find the line of intersection between
$2 x + 3y = 0$
$5x + y - z = 0 $
The point $(0,0,0)$ is an obvious solution.
The direction vector of the line is
$V = (2, 3, 0) \times (5, 1, -1) = (-3, 2, -13) $
So the line is $\ell(t) = t (-3, 2, -13)$
So now have
$ R^2 = (1 + 3 t)^2 + (1 - 2 t)^2 + (2 + 13 t)^2 = (2 + 3 t)^2 + (-2 - 2 t)^2 + (3 + 13 t)^2$
Therefore,
$1^2 + 1^2 + 2^2 + t (6 - 4 + 52) = 2^2 + (-2)^2 + 3^2 + t (12 + 8 + 78)$
So that
$ t (44) = - 11 $
and $ t = -\dfrac{1}{4} $
Therefore, the center is
$ C = -\dfrac{1}{4} (-3, 2, -13) = ( \dfrac{3}{4}, - \dfrac{1}{2} , \dfrac{13}{4} ) $
and
$ R^2 = (\dfrac{1}{4})^2 + (\dfrac{3}{2})^2 + (\dfrac{5}{4})^2 = \dfrac{62}{16} $
And the equation of the sphere is
$ (x - \dfrac{3}{4})^2 + (y + \dfrac{1}{2} )^2 + ( z - \dfrac{13}{4})^2 = \dfrac{62}{16} $
Multiplying through by $16$
$ (4 x - 3 )^2 + (4 y + 2 )^2 + (4 z - 13)^2 = 62 $
Expanding
$ 16 (x^2 + y^2 + z^2) - 24 x + 16 y - 104 z + 9 + 4 + 169 = 62 $
And finally,
$ 16 x^2 + 16 y^2 + 16 z^2 - 24 x + 16 y - 104 z + 120 = 0 $
Dividing through by $8$
$ 2 x^2 + 2 y^2 + 2 z^2 - 3 x + 2 y - 13z + 15 = 0 $