Question

The perpendicular distance from the point $(3,1,1)$ on the plane passing through the point $(1,2,3)$ and containing the line, $\overrightarrow{r}=\hat{i}+\hat{j}+\lambda (2\hat{i}+\hat{j}+4\hat{k})$, is:

• A. $\frac{1}{\sqrt{11}}$
• B. $\frac{4}{\sqrt{41}}$
• C. $0$
• D. $\frac{3}{\sqrt{11}}$

Answer 1

The correct option is C $0$

Let equation of plane through $P,Q$ & $R$ be $a(x–1)+b(y–2)+c(z–3)=0$, where

$a\hat{i}+b\hat{j}+c\hat{k}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 3\\ 2& 2& 7\end{array}\right|=\hat{i}+6j-2\hat{k}$

$\therefore (a,b,c)=(1,6,-2)$

$\therefore$ equation of plane will be

$x+6-2z=0$

$\therefore$ distance of $(3,1,1)$ from the plane, $d=\left|\frac{3+6-2-7}{\sqrt{1+36+4}}\right|=0$