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  1. The equation of the plane which is equidistant from lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{1}=\frac{z-1}{2}$ is Ax + By + Cz – 9 = 0 then A + B + C = ________.

I used the concept of parametric points as $(1+\beta ,2+2\beta,3+3\beta)$ snd $(2+3\alpha ,3+\alpha,1+2\alpha)$.The plane should be equidant from this point so midpoint will lie on the plane and also the plane is normal to the slope of the point but my solution is getting complicated

Samar Imam Zaidi
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1 Answers1

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The plane in question will be parallel to both the lines. Since we only have to find the direction ratios, we can do this by taking a cross product of the direction ratios of given lines $$\begin{array}{|c c c|} i & j & k \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{array} = i+7j-5k$$

Thus the plane should be $x + 7y-5z-9=0$. A simple test for equidistance may be done by taking two points on the two lines, eg $(1,2,3)$ and $(2,3,1)$ and calculating the distance. For both points, this distance comes out to be $\frac{3\sqrt3}{5}$. Therefore, $A+B+C = \boxed{3}$.

Aniruddha Deb
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