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Let the line \mathrm{\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}} intersect the plane containing the lines \mathrm{\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}} and \mathrm{4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a \in \mathbb{R}} at the point \mathrm{P(\alpha, \beta, \gamma)}. Then the value of \mathrm{\alpha+\beta+\gamma} equals__________.

Option: 1

12


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Equation of plane
\mathrm{4 a x-y+5 z-7 a+\lambda(2 x-5 y-z-3)=0}\text{this satisfy } (4,-1,0)
\mathrm{16 a+1-7 a+\lambda(8+5-3)=0}
\mathrm{9 a+1+10 d=0 \cdots (1)}
Normal vector of the plane A is \mathrm{(4 a+2 \lambda,-1-5 \lambda, 5-\lambda)}
vector along the line which contained the plane A is \mathrm{\hat{i}-2 j+k}
\mathrm{\therefore 4 a+2 \lambda+2+10 \lambda+5-\lambda=0}
\mathrm{11 \lambda+4 a+7=0 \cdots (a)}
Solve (1) and (2) to get \mathrm{a=1, \lambda=-1}
Now equation of plane

\mathrm{ x+2 y+3 z-2=0}

Let the point in the line \mathrm{\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}=t}
is \mathrm{(7 t+3,-t+2,-4 t+3)}  satisfy the equation of plane A 
\mathrm{7 t+3-2 t+4+9-12 t-2=0}
\mathrm{t=2}
So \mathrm{\alpha+\beta+\gamma=2 t+8=12}

Posted by

Sanket Gandhi

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