Symmetry by plane

Determine the coordinates of a image of point A (3, -4, -6) at a symmetry that is determined by the plane x-y-4z-13 = 0

Correct result:

x =  -1
y =  1
z =  4


r:x-y-4z-13=0 \ \\ A(3,-4,-6) \ \\ \ \\ n=s \ \\ n=(1,-1,-4) \ \\ p: x=t+a; y=-t+b; z=-4t + c; t \in R \ \\ \ \\ <\div class='alert alert-danger'>System of equations are non-linear.</\div> <\div class='alert alert-success'>The equations have the following integer solutions:</\div> <b>3=t + a \ \\ -4=-t + b \ \\ -6=-4 \cdot \ t + c \ \\ t=1</b><b>Number of solutions found: 1</b><h_{5}>a_{ 1</sub\geq_{2}, b_{ 1</sub\geq-3, c_{ 1</sub\geq-2, t_{ 1</sub\geq_{1}</h_{5}> \ \\ \ \\ p: x=t+2; y=-t-3; z=-4t-2; t \in R \ \\ \ \\ p(t=-1)=A' \ \\ \ \\ x=-1+a=-1
z=4 (1)+c=4z=-4 \cdot \ (-1)+c=4

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