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  • If the foot of the perpendicular drawn from (1,9,7) to the line passing through the point (3,2,1) and parallel to the planes <img alt="x + 2y + z = 0" src="https://entrancecorner.oncodeco

If the foot of the perpendicular drawn from (1,9,7) to the line passing through the point (3,2,1) and parallel to the planes

x + 2y + z = 0 and 3y - z = 3 is (\alpha, \beta, \gamma), then \alpha+\beta+\gamma is equal to 

Option: 1

3


Option: 2

1


Option: 3

-1


Option: 4

5


Answers (1)

best_answer

\begin{aligned} & \overline{\mathrm{n}}_1=(1,2,1) \overline{\mathrm{n}}_2=(0,3,-1) \\ & \overline{\mathrm{P}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & 1 \\ 0 & 3 & -1 \end{array}\right|=\hat{\mathrm{i}}(-5)-\hat{\mathrm{j}}(-1)+\hat{\mathrm{k}}(3) \\ & =(-5,1,3) \\ & \therefore \text { line }: \frac{\mathrm{x}-3}{-5}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-1}{3}=\lambda \end{aligned}

\begin{aligned} & (2-5 \lambda) \cdot-5+(\lambda-7) \cdot 1+(3 \lambda-6) \cdot 3=0 \\ & 25 \lambda-10+\lambda-7+9 \lambda-18=0 \\ & 35 \lambda=35 \\ & \lambda=1 \\ & \therefore \text { Point is }(-2,3,4) \quad \alpha+\beta+\gamma=5 \end{aligned}

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Sanket Gandhi

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