Find the intersection point if any of the lines r1t812570t10

Find the intersection point (if any) of the lines r1(t)=(81,25,70)+t(10,3,8)

and r2(s)=(189,56,20)+s(30,9,3)

r1(t)=(81,25,70)+t(10,3,8)and r2(s)=(189,56,20)+s(30,9,3)

The two lines intersect if and only if there are real numbers a, b such that:

(81,25,70)+a(10,3,8) = (189,56,20)+b(30,9,3)

So equating both sides:

81 -10a = -189+30b

25 -3a = -56 + 9b

-70 +8a = 20 -3b

on solving above we get:a =9 and b =6

So, x= 81 -10*9 = -9

y = 25 -3*9 = -2

z = -70 +8*9 = 2

Point of intersection : ( -9 , -2, 2)