# If the distance of the point

Question:

If the distance of the point $(1,-2,3)$ from the plane $x+2 y-3 z+10=0$ measured parallel to the line, $\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1}$ is $\sqrt{\frac{7}{2}}$, then the value of Iml is equal to

Solution:

DC of line $\equiv\left(\frac{3}{\sqrt{m^{2}+10}}, \frac{-m}{\sqrt{m^{2}+10}}, \frac{1}{\sqrt{m^{2}+10}}\right)$

Q lies on $x+2 y-3 z+10=0$

$1+\frac{3 \mathrm{r}}{\sqrt{\mathrm{m}^{2}+10}}-4-\frac{2 \mathrm{mr}}{\sqrt{\mathrm{m}^{2}+10}}-9-\frac{3 \mathrm{r}}{\sqrt{\mathrm{m}^{2}+10}}+10=0$

$\Rightarrow \frac{r}{\sqrt{m^{2}+10}}(3-2 m-3)=2$

$\Rightarrow \frac{r}{\sqrt{m^{2}+10}}(-2 m)=2$

$\mathrm{r}^{2} \mathrm{~m}^{2}=\mathrm{m}^{2}+10$

$\frac{7}{2} \mathrm{~m}^{2}=\mathrm{m}^{2}+10 \Rightarrow \frac{5}{2} \mathrm{~m}^{2}=10 \Rightarrow \mathrm{m}^{2}=4$

$\mathrm{Iml}=2$