The surface
$z = x^2 + y^2 \tag{1}$
is in fact the level surface of the function
$F(x, y, z) = x^2 + y^2 -z \tag{2}$
on which $F(x, y, z)$ takes the value $0$. Since the gradient of a differentiable function is normal to its level surfaces, we can indeed obtain a normal to (1) by computing the gradient of $F(x, y, z)$ as given by (2). We have
$\nabla F(x, y, z) = (2x, 2y, -1) \tag{3}$
at any point $(x, y, z) \in \Bbb R^3$; thus, at the point $(1, -2, 5)$, which is easily seen to lie in the surface
$F(x, y, z) = x^2 + y^2 - z = 0, \tag{4}$
($1^2 + (-2)^2 -5 = 0$) we have
$\nabla F(1, -2, 5) = (2, -4, -1). \tag{5}$
Now, I'm not exactly sure what it means to represent a line in vectorial form, but an expression for the line normal to the surface $z = x^2 + y^2$ at the point $(1, -2, 5)$ may now be written in terms of the normal vector $\vec n = \nabla F(1, -2, 5)$ and the position vector field $\vec r= (x, y, z)$ on $\Bbb R^3$; taking $\vec r_0 = (1, -2, 5)$, we have
$\vec r(t) = \vec r_0 + t\nabla F(1,2,5), \tag{6}$
or
$\vec r(t) = (1, -2, 5) + t(2, -4, -1). \tag{7}$
An expression for the tangent plane may be had in a roughly similar manner; $\vec r = (x, y, z)$ is a point in the tangent plane if and only if the vector $\vec r - \vec r_0$ lies in that plane and is hence perpendicular to $\nabla F(1, -2, 5)$; thus we may write
$(\vec r - \vec r_0) \cdot \nabla F(1, -2, 5) = (\vec r - \vec r_0) \cdot (2, -4, -1) =0 \tag{8}$
for the equation of the (tangent) plane passing through the point $\vec r_0 = (1, -2, 5)$. A little algebra allows (8) to be re-written in another well-known form:
$2x -4y - z = \vec r_0 \cdot (2, -4, -1) = (1, -2, 5) \cdot (2, -4, -1) = 5. \tag{9}$
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
Consider the level surface: $f(x,y,z) = x + y + z - 6e^{xyz}$. Taking gradient of $f$ at point $(0,0,6)$:
$\nabla f|_{(0,0,6)} = (f_x,f_y,f_z) = (1-6yze^{xyz},1-6xze^{xyz},1-6xye^{xyz})|_{(0,0,6)} = (1,1,1)$. This produces immediately the equation for the tangent plane $P$:
$P: 1(x-0) + 1(y-0) + 1(z-6) = 0$, or $x + y + z = 6$. For the normal line $L$, it is:
$L: (x,y,z) = (0,0,6) + t(1,1,1), t\in \mathbb{R}$.
Best Answer
Assuming all you did is correct you have to find a $t$ such that the point generated by $\mathbf r (t)$ is in the plane $x+y+2z=6$. Namely, if $t=-3$ then $\mathbf r(-3)= \langle -11,17,0\rangle$. The last point is in the plane.