a) Đ, b) Đ, c) Đ, d) S

a) \(\overrightarrow {AB} = \left( {2; - 2;0} \right)\).

b) Mặt phẳng \(\left( {Oyz} \right)\)có phương trình: \(x = 0\).

Ta có \(d\left( {A,\left( {Oyz} \right)} \right) = \frac{{\left| 1 \right|}}{{\sqrt {{1^2}} }} = 1\).

c) \(d\left( {B,\left( P \right)} \right) = \frac{{\left| {3 - 1 + 0 + 5} \right|}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {1^2}} }} = \frac{{7\sqrt 3 }}{3}\).

d) Vì \(\left( Q \right)//\left( P \right)\) nên \(\left( Q \right):x - y + z + d = 0\left( {d \ne 5} \right)\).

Vì \(d\left( {A,\left( Q \right)} \right) = d\left( {B,\left( Q \right)} \right)\) nên \(\frac{{\left| {1 - 3 + 0 + d} \right|}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {1^2}} }} = \frac{{\left| {3 - 1 + 0 + d} \right|}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {1^2}} }}\)\( \Leftrightarrow \left| {d - 2} \right| = \left| {d + 2} \right|\)

\( \Leftrightarrow \left[ \begin{array}{l}d - 2 = d + 2\\d - 2 = - d - 2\end{array} \right.\)\( \Leftrightarrow d = 0\).

Vậy \(\left( Q \right):x - y + z = 0\). Suy ra \(b = - 1;c = 1;d = 0\). Do đó \(b + c + d = 0\).