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  • Let the equation of the plane containing the line <img alt="\mathrm{x+10=\frac{8-y}{2}=z\: be \: a x+

Let the equation of the plane \mathrm{P} containing the line \mathrm{x+10=\frac{8-y}{2}=z\: be \: a x+b y+3 z=2(a+b)} and the distance of the plane \mathrm{P} from the point \mathrm{(1,27,7) be \: c}. Then \mathrm{a^2+b^2+c^2} is equal to
 

Option: 1

355


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given equation of plane is \mathrm{ a x+b y+3 z=2(a+b)}
It containing the line \mathrm{\frac{\mathrm{x}-(-10)}{1}=\frac{\mathrm{y}-8}{-2}=\frac{\mathrm{z}-0}{1}}
\therefore plane (1) must passes through (-10,8,0) and parallel to 1,-2,1 Hence,
\begin{aligned} & \quad a(-10)+8 \mathrm{~b}=2 \mathrm{a}+2 \mathrm{~b} \\ \end{aligned}

\begin{aligned} & \Rightarrow \quad 12 \mathrm{a}-6 \mathrm{~b}=0 \\ \end{aligned}................(2)

\begin{aligned} & \text { and } \quad \mathrm{a}-2 \mathrm{~b}+3=0 \\ \end{aligned}...........(3)

\begin{aligned} & \text { on solving (2) and (3), we get } \\ \end{aligned}

\begin{aligned} & \qquad \mathrm{b}=2, \mathrm{a}=1 \end{aligned}
on solving (2) and (3), we get
\mathrm{b=2, a=1}
\therefore equation of the plane is
\mathrm{x+2 y+3 z=6}    
\mathrm{c} is perpendicular distance from (1,27,7)to the plane (4)
\Rightarrow \mathrm{c}=\left|\frac{1+2 \times 27+3 \times 7-6}{\sqrt{1^2+2^2+3^2}}\right|=\left|\frac{70}{\sqrt{14}}\right|=\frac{10 \sqrt{7}}{\sqrt{2}}
Now, a^2+b^2+c^2=1+4+\frac{700}{2}=\frac{710}{2}=355

Posted by

Irshad Anwar

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