Find a plane containing the line $\bar{r}(t)=< 3,-4,-3> +t< 7,-4,6> $ and orthogonal to the plane $-4 x+2 y+2 z=-3$
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Step 1 ：The given line can be represented in parametric form as: \(x = 3 + 7t\), \(y = -4 - 4t\), \(z = -3 + 6t\).

Step 2 ：The direction vector of the line is \(<7, -4, 6>\). Any plane containing this line will have a normal vector that is orthogonal to this direction vector.

Step 3 ：The plane we are looking for is also orthogonal to the plane \(-4x + 2y + 2z = -3\). The normal vector of this plane is \(<-4, 2, 2>\).

Step 4 ：So, the normal vector of the plane we are looking for is orthogonal to both \(<7, -4, 6>\) and \(<-4, 2, 2>\). We can find this vector by taking the cross product of these two vectors.

Step 5 ：The normal vector of the plane we are looking for is \([-20, -38, -2]\). Now we can use this normal vector to find the equation of the plane. The equation of a plane given a point \((x0, y0, z0)\) on the plane and a normal vector \((a, b, c)\) is given by: \(a(x - x0) + b(y - y0) + c(z - z0) = 0\).

Step 6 ：We know that the line lies on the plane, so we can use any point on the line as a point on the plane. We can use the point \((3, -4, -3)\) for this purpose.

Step 7 ：The equation of the plane is \(-20(x - 3) + -38(y - -4) + -2(z - -3) = 0\).

Step 8 ：Simplifying the equation gives us the final answer: \(\boxed{-20x + 60 -38y - 152 -2z + 6 = 0}\) or \(\boxed{-20x -38y -2z + 86 = 0}\).