Answer
The point on the graph of $z = 3{x^2} - 4{y^2}$ at which the vector ${\bf{n}} = \left( {3,2,2} \right)$ is normal to the tangent plane is
$\left( { - \frac{1}{4},\frac{1}{8},\frac{1}{8}} \right)$
Work Step by Step
Write $f\left( {x,y} \right) = 3{x^2} - 4{y^2}$. So, the partial derivatives are
${f_x} = 6x$, ${\ \ \ }$ ${f_y} = - 8y$
Let $\left( {a,b} \right)$ be the point where the vector ${\bf{n}} = \left( {3,2,2} \right)$ is normal to the tangent plane of the graph of $f$.
By Eq. (2) the tangent plane is given by
$z = f\left( {a,b} \right) + {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right)$
$z = 3{a^2} - 4{b^2} + 6a\left( {x - a} \right) - 8b\left( {y - b} \right)$
$z = 6ax - 8by - 3{a^2} + 4{b^2}$
Or
$6ax - 8by - z = 3{a^2} - 4{b^2}$
According to Eq. (5) of Section 13.5, the normal vector to this plane is ${{\bf{n}}_1} = \left( {6a, - 8b, - 1} \right)$.
Since it is desired that ${\bf{n}} = \left( {3,2,2} \right)$ be normal to the tangent plane, ${\bf{n}}$ and ${{\bf{n}}_1}$ must be parallel. Thus, we can write ${{\bf{n}}_1} = - \frac{1}{2}\left( { - 12a,16b,2} \right)$ such that ${{\bf{n}}_1} = - \frac{1}{2}{\bf{n}}$. Therefore,
$3 = - 12a$, ${\ \ \ }$ $a = - \frac{1}{4}$
$2 = 16b$, ${\ \ \ }$ $b = \frac{1}{8}$
So, the point on the graph of $z = 3{x^2} - 4{y^2}$ at which the vector ${\bf{n}} = \left( {3,2,2} \right)$ is normal to the tangent plane is
$\left( {a,b,f\left( {a,b} \right)} \right) = \left( { - \frac{1}{4},\frac{1}{8},f\left( { - \frac{1}{4},\frac{1}{8}} \right)} \right) = \left( { - \frac{1}{4},\frac{1}{8},\frac{1}{8}} \right)$
