In a triangle ABC, right angled at the vertex A, if the position vectors of A, B, and C are respectively 3^i+^j-^k, -^i+3^j+p^k, and 5^i+q^j+^k, then the point (p,q) lies on a line:

Triangle ABC is right angled at A. So Line BA and CA are perpendicular to each other.

Let the position vectors of A, B, and C be a, b, and c respectively.
Then a = 3^i + ^j - ^k
b = -^i + 3^j + p^k
c = 5^i + q^j + ^k

The vector BA is given by b - a = (-1-3)^i + (3-1)^j + (p+1)^k = -4^i + 2^j + (p+1)^k
The vector CA is given by c - a = (5-3)^i + (q-1)^j + (1+1)^k = 2^i + (q-1)^j + 2^k

Since BA and CA are perpendicular, their dot product is zero.

BA . CA = (-4)(2) + (2)(q-1) + (p+1)(2) = 0
-8 + 2q - 2 + 2p + 2 = 0
2p + 2q - 8 = 0
p + q - 4 = 0
This is the equation of a straight line.
q = -p + 4
The slope of this line is -1. Since the slope is negative, the line makes an obtuse angle with the positive direction of the x-axis.