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If the distance of points 2 \hat{i}+3 \hat{j}+\lambda \hat{k} from the plane \bar{r}(3 \hat{i}+2 \hat{j}+6 \hat{k})=13 is 5 units then \lambda=
(A) 6,-\frac{17}{3}
(B) 6, \frac{17}{3}
(C) -6,-\frac{17}{3}
(D) -6, \frac{17}{3}
maths | Maths
If the distance of points 2 \hat{i}+3 \hat{j}+\lambda \hat{k} from the plane \bar{r}(3 \hat{i}+2 \hat{j}+6 \hat{k})=13 is 5 units then \lambda=
(A) 6,-\frac{17}{3}
(B) 6, \frac{17}{3}
(C) -6,-\frac{17}{3}
(D) -6, \frac{17}{3}

Correct option is

(A) 6,-\frac{17}{3}

Equation of plane \bar{r} \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})=13

i.e. 3 x+2 y+6 z-13=0

Given point (2,3, \lambda)

distance of plane from the points =\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

\begin{array}{l} 5=\left|\frac{3(2)+2(3)+6 \lambda-13}{\sqrt{9+4+36}}\right| \\ \therefore 5=\left|\frac{6 \lambda-1}{7}\right| \end{array}

\Rightarrow 6 \lambda-1=\pm 35

\Rightarrow 6 \lambda=36,6 \lambda=-34

\therefore \lambda=6, \lambda=-\frac{17}{3}

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