We have,
\(\bar{a_1}=-i-j-k,\bar{a_2}=3i+5j+7k,\\\bar{b_1}=7i-6j+k,\bar{b_2}=i-2j+k\\\bar{a_2}-\bar{a_1}=4i+6j+8k\)
(ii) Let the equation of the plane be ax + by + cz + d = 0….(1)
Since (1) is perpendicular to 3x + 2y + 3z = 5,
a + 2b + 3c = 0
Since (1) is perpendicular to 3x + 3y + z = 0
3a + 3b + c = 0
Using (2) and (3) we have;
(1) = -7x + 8y – 3z + d = 0
Since (1) passes through (-1, 3, 2) we have;
⇒ -7 (-1) + 8(3) – 3(2) + d = 0
⇒ 7 + 24 – 6 + d = 0 = d = -25
Therefore the equation of the plane is
(1) ⇒ -7x + 8y – 3z – 25 = 0