The equation of the plane, which bisects the line joining the points (1,2,3) and (3,4,5) at right angles is

The given points are $A(1,2,3)$ and $B(3,4,5)$ The direction ratios of line segment $A B$ is given by $$ \begin{array}{l} \left(x_{2}-x_{1}\right),\left(y_{2}-y_{1}\right),\left(z_{2}-z_{1}\right) \\ (3-1),(4-2),(5-3)=(2,2,2)=(1,1,1) \end{array} $$ Since, the plane bisects $A B$ at right angles, $\mathbf{A B}$ is the normal to the plane. Which is n Therefore, $\mathbf{n}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ Let $c$ be the mid-point of $A B$ $$ \begin{aligned} \left(\frac{x_{1}+x_{2}}{2},\right.&\left.\frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right) \\ =& c\left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{3+5}{2}\right) \\ =& c(2,3,4) \end{aligned} $$ Let this vector be $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ Hence, the required equation of vector $$ \{\mathbf{r}-(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \mathbf{k})\}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k})=0 $$ We know that, $\mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$ $$ \begin{array}{l} \Rightarrow\{(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(2\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})\}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \\ \Rightarrow \quad(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ =(2\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ \Rightarrow x+y+z=2+3+4 \\ \Rightarrow x+y+z=9 \end{array} $$