Answer
$$-x + 2y+7z+17 =0$$
Work Step by Step
Write the general equation of a plane in $R^3$ passing through the point $P=(x_0,y_0, z_0)$ along with the normal vector $n=\lt p,q,r\gt$ such that:
$$p(x - x_0) + q(y - y_0) + r(z - z_0) = 0~~~~(1)$$
Here, we have $n=\lt 1,-3, 1\gt \times \lt 4, 2, 0\gt=-2i+4j+14k$ or, $n=\lt -2, 4, 14 \gt$
Plug the values into the equation (1) to obtain:
$$ ( -2 )(x - 3) +4(y -0) + 14 (z+2) = 0 $$
After simplifying, we get:
$$-2x+4y+14z+34 =0\\ -x + 2y+7z+17 =0$$
