Answer:
-30(x-4)  + 29 (y-1)  -12(z-3)  =  0
Step-by-step explanation:
You can find the perpendicular vector to both curves using the cross product  between them, and that will give you all the information you need to find the equation of the tangent plane to the surface S. 
Let's find the parameter of each curve for the specific point. 
For the first curve  notice that 

And for the second curve notice that 

Thus for the first curve we reach our point at  and for the second curve we reach our point at
    and for the second curve we reach our point at  
 
Then we compute the tangent vector to our curves, and it would be given by the derivative so 
 and at
     and at   we have that
    we have that  

Similarly 

The the cross product between the two vectors would be 

Since (-30,29,-12) would be the perpendicular vector to the tangent plane and (4,1,3) would be a point of the tangent  then we would have that the equation of the tangent plane is given by 
-30(x-4)  + 29 (y-1)  -12(z-3)  =  0