Answer 1

Let `bara, barb,barc` be the position vectors of `triangle ABC` whose vertices are A(2, p, –3), B(q, –2, 5) and C(–5, 1, r)

`:.bara = 2hati + pbarj - 3bark`, `barb = qbari - 2barj + 5bark`, `barc = -5bari +  barj + rbark`

Given that origin O is the centroid of `triangle ABC`

`:. barO = (bara+barb+barc)/3`

`:. bara+barb+barc = barO`

`2hati + phatj - 3hatk + hatj - 2hatj +5hatk - 5hati + hatj + rhatk = barO`

`=> (2+q-5)hati + (p-2+1)hatj + (-3+5+r)hatk = 0hati + 0hatj + 0hatk`

by equality of vectors

2 + q - 5 = 0 ⇒ q = 3

p - 2 + 1 = 0 ⇒ p = 1

-3 + 5 + r = 0 ⇒ r = -2

∴ p = 1, q = 3 and r = -2