Giao điểm $A\left( {{x}_{0}};{{y}_{0}};{{z}_{0}} \right)$ của ${{d}_{1}};{{d}_{2}}$ thỏa mãn:

$\left\{ \begin{align}  & \frac{{{x}_{0}}+1}{2}=\frac{{{y}_{0}}}{3}=\frac{{{z}_{0}}-1}{3} \\  & \frac{-{{x}_{0}}+1}{2}=\frac{{{y}_{0}}}{1}=\frac{{{z}_{0}}-1}{1} \\ \end{align} \right.$

$\begin{align}  & \Rightarrow \frac{-{{x}_{0}}+1}{2}=3.\frac{{{x}_{0}}+1}{2}\Rightarrow {{x}_{0}}=-\frac{1}{2};{{y}_{0}}=\frac{3}{4};{{z}_{0}}=\frac{7}{4} \\  & \Rightarrow A\left( \frac{-1}{2};\frac{3}{4};\frac{7}{4} \right) \\  & \Rightarrow {{d}_{A/\left( P \right)}}=\frac{\left| -1+3-7-3 \right|}{\sqrt{{{2}^{2}}+{{4}^{2}}+{{4}^{2}}}}=\frac{4}{3} \\ \end{align}$