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  • Let P be the point of intersection of the line <img alt="\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx&plus;3%7D%7B3%7D%3D%5Cf

Let P be the point of intersection of the line \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}   and the plane  x + y + z = 2. If the distance of the point P from the plane 3x – 4y + 12z = 32 is q, then q and 2q are the roots of the equation :

Option: 1

x^{2}+18x-72=0


Option: 2

x^{2}+18x+72=0


Option: 3

x^{2}-18x-72=0


Option: 4

x^{2}+18x+72=0


Answers (1)

best_answer

\begin{aligned} & \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-\mathrm{z}}{2}=\lambda \\ & \mathrm{x}=3 \lambda-3, \mathrm{y}=\lambda-2, \mathrm{z}=1-2 \lambda \end{aligned}

P  \begin{aligned} (3 \lambda-3,\lambda-2,1-2 \lambda) \end{aligned}  will satisfy the equation of plane x + y + z = 2.

\begin{aligned} & 3 \lambda-3+\lambda-2+1-2 \lambda=2 \\ & 2 \lambda-4=2 \\ & \lambda=3 \\ & \mathrm{P}(6,1,-5) \end{aligned}

Perpendicular distance of P from plane 3x – 4y + 12 z – 32 = 0 is

\begin{aligned} & q=\left|\frac{3(6)-4(1)+12(-5)-32}{\sqrt{9+16+144}}\right| \\ & q=6 \\ & 2 q=12 \end{aligned}

Sum of roots = 6 + 12 = 18
Product of roots = 6(12) = 72
\therefore Quadratic equation having q and 2q as roots is  x^{2}-18+72 

 

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