Trong không gian Oxyz, đường thẳng d đi qua điểm M(3; −1; 5) và vuông góc với mặt phẳng (P): 2x + y − z + 4 = 0. Biết phương trình chính tắc của d có dạng \(\frac{{x - 3}}{a} = \frac{{y + 1}}{b} = \frac{{z - 5}}{{ - 1}}\). Tính giá trị của tổng a + b.

A. \(\left\{ \begin{array}{l}x = 1 + t\\y = 1\\z = 1\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = 1\\y = 1\\z = 1 + t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = 1 - t\\y = 1\\z = 1\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = 1\\y = 1 + t\\z = 1\end{array} \right.\).

A. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 - t\\z = 3 + 3t\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = - 1 + 2t\\y = 2 - t\\z = - 3 + 3t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = 2 + t\\y = - 1 - 2t\\z = 3 + 3t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = 1 - 2t\\y = - 2 - t\\z = 3 - 3t\end{array} \right.\).

A. \(\frac{{x + 2}}{1} = \frac{y}{{ - 2}} = \frac{z}{{ - 1}}\);

B. \(\frac{{x - 1}}{1} = \frac{{y - 2}}{2} = \frac{{z - 1}}{1}\);

C. \(\frac{{x + 1}}{1} = \frac{{y + 2}}{{ - 2}} = \frac{{z + 1}}{{ - 1}}\);

D. \(\frac{{x - 2}}{2} = \frac{y}{{ - 4}} = \frac{z}{{ - 2}}\).

A. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 + t\\z = 2 - 3t\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = 1 + t\\y = - 2 - 2t\\z = 2 + t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = 2 + t\\y = 1 - 2t\\z = - 3 + 2t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = - 1 + 2t\\y = 2 + t\\z = - 2 - 3t\end{array} \right.\).

A. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 - t\\z = 1 + t\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 - 4t\\z = 1 + 3t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = 2 + t\\y = - 1 - 2t\\z = 1 + t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = - 2 - t\\z = 1 + 3t\end{array} \right.\).

A. \({d_1}:\frac{x}{1} = \frac{{y - 1}}{{ - 1}} = \frac{z}{2}\);

B. \({d_2}:\frac{x}{1} = \frac{{y + 1}}{{ - 1}} = \frac{z}{{ - 1}}\);

C. \({d_3}:\frac{x}{1} = \frac{{y - 1}}{{ - 1}} = \frac{z}{{ - 1}}\);

D. \({d_4}:\left\{ \begin{array}{l}x = 2t\\y = 0\\z = - t\end{array} \right.\).

A. \(\left\{ \begin{array}{l}x = 2 + t\\y = - 5 + 2t\\z = 1 - t\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = 3 + 2t\\y = - 3 - 5t\\z = 1 + t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = 2 - 5t\\z = 1 + t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = 3 - 2t\\y = - 3 + 5t\\z = - t\end{array} \right.\).