Question

The line joining the points (1, 1, 2) and (3, -2, 1) meets the plane 3x + 2y + z = 6 at the point

• A. (1, 1, 2)
• B. (3, -2, 1)
• C. (2, -3, 1)
• D. (3, 2, 1)

Answer 1

The correct option is B (3, -2, 1)

The straight line joining the points (1, 1, 2) and (3, -2, 1) is $\frac{x-1}{2}=\frac{y-1}{-3}=\frac{z-2}{-1}=r$ (say)
$\therefore$ Point is (2r + 1, 1 -3r, 2 –r )
Which lies on 3x + 2y + z = 6
$\therefore 3(2r+1)+2(1-3r)+2-r=6$
$\therefore r=1$
Required point is (3, -2, 1).