A vector in the direction of the required line can be obtained by cross product of

$$\begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{array}\right| \\\\ & =-9 \hat{i}-9 \hat{j}+9 \hat{k} \end{aligned}$$

Required line

$$\begin{aligned} & \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda^{\prime}(-9 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}) \\\\ & \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \end{aligned}$$

Now distance of $$(0,2,-2)$$

$$\begin{aligned} & \text { P.V. of } \mathrm{P} \equiv(5+\lambda) \hat{i}+(\lambda-4) \hat{j}+(3-\lambda) \hat{k} \\\\ & \overrightarrow{\mathrm{AP}}=(5+\lambda) \hat{i}+(\lambda-6) \hat{j}+(5-\lambda) \hat{k} \\\\ & \overrightarrow{\mathrm{AP}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\\\ & 5+\lambda+\lambda-6-5+\lambda=0 \\\\ & \lambda=2 \\\\ & |\overrightarrow{\mathrm{AP}}|=\sqrt{49+16+9} \\\\ & |\overrightarrow{\mathrm{AP}}|=\sqrt{74} \end{aligned}$$