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)#.