Gọi \[M\left( {a\,;0\,;0} \right),N\left( {0\,;b\,;0} \right),P\left( {0\,;0\,;\,c} \right)\].

\[\begin{array}{l}\overrightarrow {DM} = \left( {a - 4;1; - 3} \right)\\\overrightarrow {DN} = \left( { - 4;b + 1; - 3} \right)\\\overrightarrow {DP} = \left( { - 4;1;c - 3} \right)\end{array}\]

Ta có \[DM,DN,DP\] đôi một vuông góc với nhau nên

\[\left\{ \begin{array}{l}\overrightarrow {DM} .\overrightarrow {DN} = 0\\\overrightarrow {DM} .\overrightarrow {DP} = 0\\\overrightarrow {DN} .\overrightarrow {DP} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 4\left( {a - 4} \right) + b + 1 + 9 = 0\\ - 4\left( {a - 4} \right) + 1 - 3\left( {c - 3} \right) = 0\\16 + b + 1 - 3\left( {c - 3} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 4a + b = - 26\\ - 4a - 3c = - 26\\b - 3c = - 26\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{{13}}{4}\\b = - 13\\c = \frac{{13}}{3}\end{array} \right.\].

a) Sai.

b) Sai.

c) Đúng vì: \[{V_{DMNP}} = \frac{1}{6}DM.DN.DP = \frac{1}{6}.\frac{{13}}{4}.13.\frac{{13}}{3} = \frac{{2197}}{{72}} > 29\].

d) Gọi \[\overrightarrow x = \left( {m\,;n\,;p} \right)\]

\[\overrightarrow {DM} = \left( { - \frac{3}{4};\,1\,; - 3} \right);\,\,\overrightarrow {DN} = \left( { - 4; - 12; - 3} \right);\,\,\,\overrightarrow {DP} = \left( { - 4;1\,;\,\frac{4}{3}} \right)\]

\[\left\{ \begin{array}{l}\overrightarrow x .\overrightarrow {DM} = 1\\\overrightarrow x .\overrightarrow {DN} = 2\\\overrightarrow x .\overrightarrow {DP} = - 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - \frac{3}{4}m + n - 3p = 1\\ - 4m - 12n - 3p = 2\\ - 4m + n + \frac{4}{3}p = - 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = \frac{{88}}{{169}}\\n = - \frac{{35}}{{169}}\\p = - \frac{{90}}{{169}}\end{array} \right.\]

\[m\, + n\, + p = \frac{{ - 37}}{{169}}\].

Suy ra d) sai.

Trong không gian Oxyz , cho vectơ b = ( 2 ; − 2 ; 1 ) . Tìm giá trị m sao cho m vectơ b = vectơ 0 .