Let {{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}} lie on the plane px - qy + z = 5, for some p, q \in R. The shortest distance of the plane from the origin is :

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Answer: B. \sqrt {{5 \over {142}}} \frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}=\lambda (3 \lambda+2,-2 \lambda-1,-\lambda-3) lies on plane p x-q y+z=5 p(3 \lambda+2)-q(-2 \lambda-1)+(-\lambda-3)=5 \lambda(3 p+2 q-1)+(2 p+q-8)=0 3 p+2 q-1=0\} p=15 2 p+q-8=0\} q=-22 Equation of plane 15 x+22 y+z-5=0 Shortest distance from origin =\frac{|0+0+0-5|}{\sqrt{15^{2}+22^{2}+1}} =\frac{5}{\sqrt{710}} =\sqrt{\frac{5}{142}}

Let $${{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}}$$ lie on the plane $$px - qy + z = 5$$, for some p, q $$\in$$ R. The shortest distance of the plane from the origin is :

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