Question

The plane passing through the point $(-2,-2,2)$ and containing the line joining the points $(1,1,1)$ and $(1,-1,2)$ makes intercepts on the coordinate axes, then sum of the lengths of intercepts is

• A. $3$
• B. $4$
• C. $6$
• D. $12$

Answer 1

The correct option is D $12$

normal vector would be perpendicular to both $\overline{AC}$ and $\overline{BC}$

$\Rightarrow \overrightarrow{n}=\overline{AC}\times \overline{BC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 0\\ 0& -2& 1\end{array}\right|$

$=\hat{i}-3\hat{j}-6\hat{k}$

$\Rightarrow$ Equation of plane : $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0\Rightarrow \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$

$\Rightarrow \overrightarrow{r}.(\hat{i}-3\hat{j}-6\hat{k})=(\hat{i}+\hat{j}+\hat{k}).(\hat{i}-3\hat{j}-6\hat{k})$

$\Rightarrow x-3y-6z=-3-6\Rightarrow x-3y-6z=-8$

$\Rightarrow \frac{x}{(-8)}+\frac{y}{\left(8/3\right)}+\frac{z}{\left(4/3\right)}=1$

$\Rightarrow$ sum of intercepts of lengths of$=8+\frac{2}{3}+\frac{4}{3}=\frac{36}{3}=12\Rightarrow \left(D\right)$