I need to calculate the distance of the point $P = (0, 5, -4)$ from the plane which pass from the point $P1=0, 1, -2)$ and generated by the two vectors: $$ v1 = (1, 2, 3), v2 = (-1, \sqrt{2}, 1) $$
Here is my solution but I am not sure it is the right one.
First of all I need the cartesian equation of the plane ($\pi$ from now). I know that the generic vector $Px = (x, y, z)$ is on the plane $\pi$ if the vector $Px - P1$ is coplanar with vectors $v1$ and $v2$.
So I can find the cartesian equation of the plane $\pi$ computing the determinant of the following matrix (which by definition of coplanarity is equal to 0): $$ det\begin{bmatrix}x & y-1 & z+2 \\1 & 2 & 3 \\-1 & \sqrt{2} & 1\end{bmatrix} = 0 $$
The cartesian equation is: $$ (2-3\sqrt{2})x-4y+(2+\sqrt{2})z+8+2\sqrt{2}=0 $$
Now I can apply the formula to get the distance between a point $(x0, y0, z0)$ and a plane ($ax+by+cz+d=0$), which is:
$$ \delta = \frac{|ax0+by0+cz0+d|}{\sqrt{a^2+b^2+c^2}} $$
In our case we have: $$ \delta = \frac{|(2-3\sqrt{2})*0-4*5-4*(2+\sqrt{2})+8+2\sqrt{2}|}{\sqrt{(2-3\sqrt{2})^2 + 16 + (2+\sqrt{2})^2}}= \frac{20+2\sqrt{2}}{\sqrt{44-8\sqrt{2}}} $$
Since I don't know how to validate my solution, can you plase give a feedback?
this is only a hint.
here is another way to the problem. so any point on the plane has the parametric representation $m=(0,1,-2)+s(-1,\sqrt2,1) + t(1,2,3)=a+su+tv$ and let this point be the foot of the perpendicular drawn from the point $p=(0, 5, -4).$ $t,s$ will be determined by making $\vec{pm} = m-p$ orthogonal to both vectors $u,v.$
solve $$\pmatrix{u.u & u.v\\u.v &v.v}\pmatrix{s\\t} = \pmatrix{(p-a).v\\(p-a).v}$$ for $t,s.$ that will give the point $m$ then compute $|p-m|$ that ought to give you a way to check your answer.