766. On the sphere $x^{2}+y^{2}+z^{2}=676$ find the points where the tangent plane is parallel to the plane $3 x-12 y+4 z=0$.
Solution. Using the general equation (I), we will form the equation of the tangent plane to the given sphere at its point $\left(x_{0}, y_{0}, z_{0}\right)$:

$$
x_{0}\left(x-x_{0}\right)+y_{0}\left(y-y_{0}\right)+z_{0}\left(z-z_{0}\right)=0
$$

$$
x_{6} x+y_{0} y+z_{0} z=x_{0}^{2}+y_{0}^{2}+z_{j}^{4}=676
$$

According to the condition of parallelism of two planes, in order for the tangent plane to be parallel to the given plane, the coefficients of the current coordinates in their equations must be proportional: $\frac{x_{0}}{3}=\frac{y_{0}}{-1 \overline{2}}=\frac{z_{0}}{4}=\lambda$.

From this, we determine $x_{0}=3 \lambda, y_{0}=-12 \lambda, z_{0}=4 \lambda$ and substituting into the equation of the sphere, we find two values of the proportionality coefficient: $\lambda= \pm 2$ and two desired points on the sphere $(6 ;-24$; 8) and ( $-6 ; 24 ;-8$ ), at which the tangent plane is parallel to the given plane.