With #p = (x,y,z)# and calling #C(p)-> x^2+2y^2+z^2-4=0#
The support line to the tangent planes to #C# is determined as follows.
Taking the three planes
#{(Pi_1-> x + y + z + 1=0),(Pi_2->2 x + 3 y + 2 z - 3=0 ),(Pi_g->a x+b y+c z+d=0):}#
and solving for #x,y,z# we obtain
#{(x = -(5 b - 6 c + d)/(a - c)),(y=5),(z=-6 + (5 b - 6 c + d)/(a - c)):}#
now calling #lambda = (5 b - 6 c + d)/(a - c) # we obtain the support line equation
#L->p = p_0 + lambda vec v#
with
#p_0 = (0,5,-6)#
#vec v = (-1,0,1)#
The tangent planes have the equation
#Pi_t -> << vec n_t, p-p_t >> = 0#
where
#vec n_t = grad C(p_t) = (2x_t,4y_t,2z_t)# and
#p_t = (x_t,y_t,z_t)# is the tangency point.
The tangency point #p_t# is found solving
#{(Pi_t @ L -> << vec n_t, p_0+lambda vec v-p_0 >> = lambda << vec n_t, vec v >> = 0),(C(p_t)=0),( << vec n_t, p_t-p_0 >> = 0):}#
or
#{(z_t-x_t=0),(x_t^2 + 2 y_t^2 + z_t^2 -4=0),(2 x_t^2 + 4 (y_t-5) y_t + 2 z_t (6 + z_t) =0):}#
Solving we get at two tangency points
#p_t^1 = ( -23/17, -7/17, -23/17)# and
#p_t^2=(1,1,1)#
and consequently
#vec n_t^1 = (-46/17, -28/17, -46/17)#
#vec n_t^2 = (2, 4, 2)#
and the tangent planes are
#Pi_t^1->8 + (46 x)/17 + (28 y)/17 + (46 z)/17 = 0#
#Pi_t^2-> 2 x + 4 y + 2 z-8 = 0#
