In the $xyz$ -coordinate system, a massive particle is initially at rest on a plane whose equation is given below: $x + 2y + 3z = 0.$ The particle is launched with initial velocity $(v_x,v_y,v_z) = (-3,7,12)$ . It travels under the influence of a gravitational acceleration of $(a_x,a_y,a_z) = (0,0,-10)$ until it intersects the plane once more.

When the particle intersects the plane for the second time, how far away is it from the launch point (to 3 decimal places)?

The answer is 26.484.

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Let the particle be thrown from the origin. Since there is no acceleration in the $x$ and the $y$ directions, therefore, the $x$ and $y$ coordinate of the particle after time $t$ will be, $x= -3t, \,\,\, y=7t .$ There is an acceleration of -10 units in the $z$ direction. Therefore, using the equation of motion we may write the $z$ coordinate as $z=12t - \frac{1}{2} 10 t^2 .$

When the particle lands back on the plane, these coordinates will satisfy the equation of the plane. Hence, $-3t +14t +36t -15t^2 = 0 \,\,\,\,\Rightarrow t = \frac{47}{15}.$

After time $t$ , the distance of the particle from the origin (the launch point) is $\begin{aligned} d &= \sqrt{x^2 + y^2 +z^2} \\ &= \sqrt{(-3t)^2 + (7t)^2 + \left(12t - \frac{1}{2} \cdot 10 t^2\right)^2 } . \end{aligned}$

Substituting the value of $t$ , we will get $d=\boxed{26.484}$ .