Problem:
The plane containing the point $( 1, -5, 8)$ and orthogonal to the line given by these three equations: \begin{align*} x &= -3 + 15t \\ y &= 14 - t \\ z &= 9 - 3t \end{align*}
Answer:
One of the general forms of a plane is: $$ A(x-x_0) + B(y - y_0) + C(z - z_0) = 0 $$ In this case, we have: \begin{align*} A &= 15 \\ B &= -1 \\ C &= -3 \\ 15( x - 1) + -1(y + 5) + -3(z - 8) &= 0 \\ 15x - 15 - y - 5 - 3z + 24 &= 0 \\ 15x - y - 3z + 4 &= 0 \end{align*}
Is my solution correct?