Question 1048763
A normal vector to the plane x-3y+5z=-1 is < 1,-3, 5 >.

Since the points (-4,-7,7) and (-2,-1,-6) are on the desired plane, a vector parallel to the desired plane is < -2--4, -1--7, -6-7 > = < 2,6, -13 >.


Let < a,b,c > be the normal vector of the desired plane.

< 2,6, -13 >*< a,b,c > = 0, or 

2a + 6b - 13c = 0  <------Equation A


Also,  < 1,-3,5 >*< a,b,c > = 0, or 

a - 3b + 5c = 0,  <-------Equation B

as per the given that the two planes are perpendicular.


Multiplying Equation B by 2 and adding with Equation A gives 

{{{a = (3/4)c}}}.

Multiplying Equation B by -2 and adding with Equation A gives 

{{{b = (23/12)c}}}.

===> The normal vector < a,b,c > = c < 3/4, 23/12,1 >.

Choosing c = 12, let the normal be < 9, 23, 12 >.


Therefore, the desired plane has the equation


9(x+2)+23(y+1)+12(z+6) = 0.


I leave the simplifying to you.