In what way did you arrive at that canonical form? It sounds to me as though you had essentially diagonalized the matrix of the quadratic form. You can do that, but doing so corresponds to a projective transformation so it will change the relationship between the affine space and the quadric. Therefore I'd suggest not doing this here.
Instead of merely going for $x\neq 0$ you might as well pick one specific representative and assume $x=1$. So you are looking for points $(y,w,z)\in\mathbb A^3$ with $y+yw-w^2=0$. This is independent of $z$, so you can plot this as a planar curve and then extrude it along the $z$ axis. To graph the planar curve, a hyperbola, you can compute asymptotes and probably a few values, then judge a reasonable interpolation in between. Unless you have learned different techniques.
You can do this without finding an intersection. Just find normal vectors to both surfaces (i.e. compute gradients) and then compute theirs cross product to obtain curve tangent vector.
For the first surface the gradient is
$$
(4x, 4y, -2z).
$$
The second one gives us
$$
(2x, 2x, -2z).
$$
So, we can take $(2x, 2y, -z)$ and $(x, y, -z)$ as normal vectors and compute their cross product to obtain tangent vector
$$
\boldsymbol{v} = \begin{vmatrix}
\boldsymbol{i} & \boldsymbol{j} &\boldsymbol{k} \\
2x & 2y & -z \\
x & y & -z
\end{vmatrix} = -yz \boldsymbol{i} +xz\boldsymbol{j}.
$$
Let's substitute $(3,4,5)$
$$
\boldsymbol{v} = -4\cdot 5\boldsymbol{i} + 3\cdot{5}\boldsymbol{j}.
$$
It's easy to see that normalizing this vector we get
$$
\boldsymbol{v}_1 = \frac{\boldsymbol{v}}{|\boldsymbol{v}|} = -\frac{4}{5} \boldsymbol{i} + \frac{3}{5}\boldsymbol{j}.
$$
Now you need to find the directional derivative along $\boldsymbol{v}_1$ or $-\boldsymbol{v}_1$ (both will be tangent vectors indicating two different directions on the curve).
The other way for solving this problem (as suggested in the comments) was to eliminate $z$ from the equations and get the curve (WLOG we take positive $z$ considering the case $z = -5\sqrt{2}$ by analogy)
$$
x^2 + y^2 = 50, \; z = 5\sqrt{2}.
$$
So, we see that the curve is a circle $x^2 + y^2 = 50$ located at the "height" $z=5\sqrt{2}$ (this actually explains why $k$ coordinate is equal to zero in tangent vector). The tangent vector to the circle
$$
x^2 + y^2 = 50
$$
is equal to $(-y, x)$ and after substitution we get again
$$
\boldsymbol{v} = (-4, 5) \implies \boldsymbol{v}_1 = \frac{\boldsymbol{v}}{|\boldsymbol{v}|} = -\frac{4}{5} \boldsymbol{i} + \frac{3}{5}\boldsymbol{j}.
$$
All you need now is to compute the directional derivative (i.e. to find dot product of gradient of the function and obtained unit vector).
Best Answer
Let $x=5\cos t,y=3\sin t$ then $z=4\sin t$ and so $$r(t)=(5\cos t,3\sin t,4\sin t)\implies t=3\pi/2\iff r=(0,-3,-4) \\ r'(t)=(-5\sin t,3\cos t,4\cos t) $$ the parallel to the line is $u=(5,0,0)$ and the equatio of $L$ is $r_1(t)=(0,-3,-4)+t(5,0,0)\implies x=5t,y=-3,z=-4 $. Finally the point of intersection is $$2(5t)-3(-3)-4(-4)=27\implies t=1/5\implies point=(1,-3,-4)$$