it's given the plane determined 2 equations of lines l1 and l2 and it is given point P(3,-4,2) outside the plane find the coordinate of P1 symetric with the plane.

$\displaystyle l_1 : \frac{x-5}{13}=\frac{y-6}{1}=\frac{z+3}{-4}$

$\displaystyle l_2 : \frac{x-2}{13}=\frac{y-3}{1}=\frac{z+3}{-4}$

what i did so far is :

I've found the normal vector of the plane using one vector of the lines and vector M1M2 . M1 is point of line 1 and m2 of l2.

then i found the equation of the plane using this vector and and a point M1 or M2.

then i found the equation of the normal line from point P and the normal vector of the plane.

am i right till here

Now how to find the coordinates of P1??

have i to find the length from P to plane, and the point witch the normal line intersect the plane and then using the same distance i should find P1 from point K( where the normal line intersect the plane).

normal = perpendicular