Finding the perpendicular distance of a point from a plane determined by 3 points I came across a problem which is as follows-
Let
$$\frac{x}{2}=\frac{y}{-1}=\frac{z}{3}\\
\frac{x}{-1}=\frac{y}{3}=\frac{z}{\frac{5}{3}}\\
\frac{x}{-32}=\frac{y}{-19}=\frac{z}{15}$$
be three lines. A plane intersecting these lines at $A$, $B$ and $C$ respectively such that $PA = 2$, $PB = 3$ and $PC = 6$ ( $P$ is the origin) Find the perpendicular distance of the plane from the point $P$.
My attempt: I assumed general points on the lines as $(2a,-1a,3a)$ , $(-b,3b,5/3b)$, $(-32c,-19c,15c)$ respectively
and using the given distance of the points from the the origin, the values of $a$, $b$ and $c$ come out to be $2/(14)^{1/2}, 9/(115)^{1/2}, 6/(1610)^{1/2} $ respectively. But I have no idea how to proceed further.
Any help would be greatly appreciated!
 A: We have the lines
$$
\cases{
L_1\to p_1 = \lambda_1\vec v_1\\
L_2\to p_2 = \lambda_2\vec v_2\\
L_3\to p_3 = \lambda_3\vec v_3\\
}
$$
and the plane $\Pi\to p\cdot\vec n = c$. Assuming $\vec v_k,\vec n$ normalized we have
$$
\lambda_k\vec v_k\cdot\vec n = c\Rightarrow \lambda_k = \frac{c}{\vec v_k\cdot\vec n}
$$
and then we have to solve
$$
\cases{
\lambda_k = \frac{c}{\vec v_k\cdot\vec n},\ \ k=1,2,3\\
\|\vec n\|=1
}
$$
four equations and four unknowns,($\vec n, c$)
NOTE
Here
$$
\cases{
\vec v_1 = (2,-1,3)/\sqrt{2^2+1+3^2}\\
\vec v_2 = (-1,3,\frac 53)/\sqrt{\cdots}\\
\vec v_3 = (-32,-19,15)/\sqrt{\cdots}\\
\lambda_1 = 2\\
\lambda_2 = 3\\
\lambda_3 = 6
}
$$
once determined the plane $p\cdot\vec n=c$ we have
$$
d \vec n\cdot\vec n = c\Rightarrow d = c
$$
Attached a plot shoving the plane and in red the distance, all according to the calculations.

A: Hint:
Using geometry, we'll find the distance $d$ from $P$ to the plane $ABC$.
The volume of the tetrahedron $PABC$ is
$$\frac{1}{3} \cdot d \cdot \operatorname{Area} ABC$$
Therefore, the distance $d$ is
$$d = \frac{3 \operatorname{Vol}PABC}{\operatorname{Area} ABC}$$
Now, the volume of the tetrahedron equals
$$\operatorname{Vol}PABC = \frac{1}{6} |\det (PA, PB, PC)|$$
while
$$\operatorname{Area} ABC = \frac{1}{2} \| AB \times AC \|$$
Putting all together we get
$$d = \frac{ |\det (PA, PB, PC)|}{\| AB \times AC \|}$$
