iu 9. Cho bốn điểm A(1;-1;5),B(0;0;1),C(0;2;1),D(0;3;1) . Mặt phẳng (P) chứa A,B và song song với đường thẳng CD có phương trình là 4x-z+1=0 B. 4x+y-z+1=0 2x+z-5=0 D. x+4z-1=0

Let A = (1, -1, 5), B = (0, 0, 1), C = (0, 2, 1), D = (0, 3, 1).<br />The vector $\vec{CD} = D - C = (0, 1, 0)$.<br />The vector $\vec{AB} = B - A = (-1, 1, -4)$.<br /><br />The plane (P) contains A and B, and is parallel to the line CD. Therefore, the normal vector of the plane (P) is perpendicular to both $\vec{AB}$ and $\vec{CD}$. We can find the normal vector by taking the cross product of $\vec{AB}$ and $\vec{CD}$:<br /><br />$\vec{n} = \vec{AB} \times \vec{CD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & -4 \\ 0 & 1 & 0 \end{vmatrix} = (4)\mathbf{i} - (0)\mathbf{j} + (-1)\mathbf{k} = (4, 0, -1)$<br /><br />The equation of the plane (P) is given by:<br />$4(x - x_0) + 0(y - y_0) - 1(z - z_0) = 0$<br />where $(x_0, y_0, z_0)$ is a point on the plane. We can use point A(1, -1, 5):<br />$4(x - 1) - (z - 5) = 0$<br />$4x - 4 - z + 5 = 0$<br />$4x - z + 1 = 0$<br /><br />Therefore, the equation of the plane (P) is $4x - z + 1 = 0$.<br /><br />Final Answer: The final answer is $\boxed{A}$