Find the equation of a plane through the point \( (6,-7,-8) \) and orthogonal to the line \( \left\{\begin{array}{l}x(t)=-1+8 t \\ y(t)=6+2 t \\ z(t)=8+8 t\end{array}\right. \)

To find the equation of the plane through the point \( (6,-7,-8) \) and orthogonal to the given line, we first need to determine the direction vector of the line from its parametric equations. The direction vector can be extracted from the coefficients of \( t \) in the equations: - \( x(t) = -1 + 8t \) indicates a change in the \( x \)-direction of \( 8 \). - \( y(t) = 6 + 2t \) indicates a change in the \( y \)-direction of \( 2 \). - \( z(t) = 8 + 8t \) indicates a change in the \( z \)-direction of \( 8 \). Thus, the direction vector of the line is \( \vec{d} = \langle 8, 2, 8 \rangle \). Since the plane is orthogonal to the line, the normal vector \( \vec{n} \) of the plane is parallel to this direction vector. This means \( \vec{n} = \langle 8, 2, 8 \rangle \). The general equation of a plane with a normal vector \( \langle A, B, C \rangle \) passing through a point \( (x_0, y_0, z_0) \) is given by: \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \] Substituting \( A = 8 \), \( B = 2 \), \( C = 8 \), and the point \( (6, -7, -8) \): \[ 8(x - 6) + 2(y + 7) + 8(z + 8) = 0 \] Expanding this, we have: \[ 8x - 48 + 2y + 14 + 8z + 64 = 0 \] Combining like terms, the equation becomes: \[ 8x + 2y + 8z + 30 = 0 \] To write it in a more standard form, we can rearrange it: \[ 8x + 2y + 8z = -30 \] Thus, the equation of the plane is: \[ 8x + 2y + 8z + 30 = 0 \]