The answer is: #(3,-2,4)#.
First of all let's write the line in parametric form.
If we put #x=t#, than:
#t/3=(y+7)/5rArry=-7+5/3t#
and:
#t/3=(z-2)/2rArrz=2+2/3t#.
So the line is:
#x=t#
#y=-7+5/3t#
#z=2+2/3t#
the direction of this line is #(1,5/3,2/3)#.
To write a plane, given a point #P(x_p,y_p,z_p)# and the direction #vecv(a,b,c)# of a perpendicular at the plane itself, we can use this formula:
#a(x-x_p)+b(y-y_p)+c(z-z_p)=0#,
so:
#1(x-2)+5/3(y+1)+2/3(z-3)=0rArr#
#3x-6+5y+5+2z-6=0rArr3x+5y+2z+7=0#.
Now, to find the point, it is necessary to make a system with the plane and the given line:
#3x+5y+2z+7=0#
#x=t#
#y=-7+5/3t#
#z=2+2/3t#
#rArr3t-35+25/3t+4+4/3t-7=0rArr38/3t=38rArrt=3#.
Now let's substitue this value of #t# in the three equation of the line:
#x=3#
#y=-7+5/3*3=-2#
#z=2+2/3*3=4#.
The point is: #(3,-2,4)#.