Question

# Let the line $$\displaystyle \frac{x-2}{3}= \frac{y-1}{-5}= \frac{z+2}{2}$$ lie in the plane $$x+3y-\alpha z+\beta = 0$$. Then $$\left ( \alpha ,\beta \right )$$ equals :

A
(6,7)
B
(5,15)
C
(5,5)
D
(6,17)

Solution

## The correct option is A $$\left ( -6,7 \right )$$The line is $$\displaystyle \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}2{}$$The direction ratios of the line are $$(3,-5,2)$$As the line is in the plane $$x+3y-az+ \beta =0$$,We have $$\left ( 3 \right )\left ( 1 \right )+\left ( -5 \right )\left ( 3 \right )+2\left ( -\alpha \right )=0$$$$\Rightarrow-12-2 \alpha =0$$$$\therefore \alpha = -6$$Again $$(2,1,-2)$$ lies on the plane$$\Rightarrow 2+3+2 \alpha + \beta =0$$$$\Rightarrow \beta = -2 \alpha -5=12-5=7$$Hence, $$\left ( \alpha ,\beta \right )$$ is $$\left ( -6,7 \right )$$Maths

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