Let $\lambda_{1}, \lambda_{2}$ be the values of $\lambda$ for which the points $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2,0,1)$ are at equal distance from the plane $2 x+3 y-6 z+7=0$. If $\lambda_{1} > \lambda_{2}$, then the distance of the point $\left(\lambda_{1}-\lambda_{2}, \lambda_{2}, \lambda_{1}\right)$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$ is ____________.

Since $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2,0,1)$ are equidistant <br/><br/>from plane $2 x+3 y-6 z+7=0$ <br/><br/>$$ \begin{aligned} & \therefore\left|\frac{2\left(\frac{5}{2}\right)+3(1)-6(\lambda)+7}{\sqrt{2^2+3^2+6^2}}\right|=\left|\frac{2(-2)+3(0)-6(1)+7}{\sqrt{2^2+3^2+6^2}}\right| \\\\ & \Rightarrow|5+3-6 \lambda+7|=|-4-6+7| \\\\ & \Rightarrow|15-6 \lambda|=|-3| \\\\ & \Rightarrow 15-6 \lambda= \pm 3 \\\\ & \Rightarrow 15-6 \lambda=3 \text { or } 15-6 \lambda=-3 \\\\ & \Rightarrow 6 \lambda=12 \text { or } 6 \lambda=18 \\\\ & \Rightarrow \lambda=2 \text { or } \lambda=3 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \because \lambda_1>\lambda_2 \\\\ & \therefore \lambda_1=3 \text { and } \lambda_2=2 \end{aligned} $$ <br/><br/>So, point will be $(1,2,3)$ <br/><br/>Let $\mathrm{M}_0=(1,2,3)$ <br/><br/>$M_1$ is the point through which line passes i.e, $(5,1,-7)$ <br/><br/>and $\vec{s}=\hat{i}+2 \hat{j}+2 \hat{k}$ <br/><br/>$$ \therefore \overrightarrow{\mathrm{M}_0 \mathrm{M}_1}=4 \hat{i}-\hat{j}-10 \hat{k} $$ <br/><br/>Now, required distance $=\left|\frac{\overrightarrow{\mathrm{M}_0 \mathrm{M}_1} \times \vec{s}}{|\vec{s}|}\right|$ <br/><br/>$$ \begin{aligned} & =\frac{|(4 \hat{i}-\hat{j}-10 \hat{k}) \times(\hat{i}+2 \hat{j}+2 \hat{k})|}{\sqrt{1+4+4}} \\\\ & =\frac{|18 \hat{i}-18 \hat{j}+9 \hat{k}|}{3}=9 \end{aligned} $$