Let there be a column of square cross-section having side length 2 units and of infinite length centred on z-axis. Let P be a random point lying in the section cut off by the plane  $x+2y+2z=30$  in the column. Q is reflection of P in the plane  $z=15$  and R is locus of point Q, then 

Locus of Point Q

Given:

• Square column: –1 ≤ x ≤ 1, –1 ≤ y ≤ 1, length infinite in z
• P lies in the region cut by plane: x + 2y + 2z = 30
• Q is reflection of P in plane z = 15

Solution Steps:

1. Reflection formula: Q = (x, y, 30 – z)
2. Substitute into plane equation for P: x + 2y + 2(30 – z) = 30
3. Simplified locus for Q: x + 2y – 2z = –30
4. With constraints: –1 ≤ x ≤ 1, –1 ≤ y ≤ 1

Locus Equation (Final Answer):

x + 2y – 2z = –30 
–1 ≤ x ≤ 1, –1 ≤ y ≤ 1