diketahui titik-titk P(3,3,6),Q(4,1,10) ,dan R(1,x,y) terletak dalam satu garis. nilai x+y=

PQ = kQR
Q-P = k(R-Q)
(4,1,10) - (3,3,6) = k((1,x,y) - (4,1,10)
(1,-2,4) = k(-3, x-1, y-10)

1 = k(-3)
k = -1/3

-2 = k(x - 1)
-2 = -1/3 (x - 1)
6 = x - 1
x = 6+1
x = 7

4 = k(y - 10)
4 = -1/3 (y - 10)
-12 = y - 10
y = 10 - 12
y = -2

x + y = 7 + (-2) = 5