The co-ordinates of the point at which the line $\frac{x−3}{3} ​=\frac{y+1}{2}​ =\frac{z−4}{-2}$ ​ crosses xy plane, are

(9, 3, 0)

The correct answer is Option (3) → (9, 3, 0)

Given: Line in symmetric form:

$\frac{x - 3}{3} = \frac{y + 1}{2} = \frac{z - 4}{-2} = \lambda$

To find: Point where the line crosses the xy-plane (i.e., $z = 0$)

From: $\frac{z - 4}{-2} = \lambda$ ⟹ $z = -2\lambda + 4$

Set $z = 0$:

$0 = -2\lambda + 4$ ⟹ $\lambda = 2$

Now, using $\lambda = 2$:

$x = 3\lambda + 3 = 3(2) + 3 = 9$

$y = 2\lambda - 1 = 2(2) - 1 = 3$

$z = 0$

$(9, 3, 0)$