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Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the point $$(2,-5,11)$$, then $$\mathrm{d}^{2}$$ is equal to :

JEE · Math · previous-year question

  1. A.$$\frac{147}{2}$$
  2. B.96
  3. C.$$\frac{32}{3}$$correct
  4. D.54

Answer

C. $$\frac{32}{3}$$

Explanation

Let $\langle a, b, c\rangle$ be direction ratios of plane containing $$ \begin{aligned} &\text { lines } \frac{x}{2}=\frac{y}{3}=\frac{z}{5} \text { and } \frac{x}{3}=\frac{y}{7}=\frac{z}{8} . \\\\ &\therefore \quad 2 a+3 b+5 c=0 \quad \ldots \text { (i) } \\\\ &\text { and } 3 a+7 b+8 c=0 \quad \ldots \text { (ii) } \\\\ &\text { from eq. (i) and (ii) }: \frac{a}{24-35}=\frac{b}{15-16}=\frac{c}{14-9} \\\\ &\therefore \text { D. R's} \text {. of plane are }<11,1,-5> \end{aligned} $$ Let $D . R$'s of plane $P$ be $$ then. $11 a_{1}+b_{1}-5 c_{1}=0$ and $9 a_{1}-b_{1}-5 c_{1}=0$ From eq. (iii) and (iv) : $$ &\frac{a_{1}}{-5-5}=\frac{b_{1}}{-45+55}=\frac{c_{1}}{-11-9} &\therefore D.R's of plane P are $$ Equation plane $P$ is : $1(x-3)-1(y+4)+2(z-7)=0$ $$ \Rightarrow x-y+2 z-21=0 $$ Distance from point $(2,-5,11)$ is $d=\frac{|2+5+22-2|}{\sqrt{6}}$ $\therefore d^{2}=\frac{32}{3}$

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