$${P_1}:x - y + 2z = 2$$

$${P_2}:2x + y - 3 = 2$$

JEE Main 2021 (Online) 22th July Evening Shift Mathematics - 3D Geometry Question 250 English Explanation

Let line of Intersection of planes P₁ and P₂ cuts xy plane in point Q.

$$\Rightarrow$$ z-coordinate of point Q is zero

$$ \Rightarrow \left. {\matrix{ {x - y = 2} \cr {and\,2x + y = 2} \cr } } \right\} \Rightarrow x = {4 \over 3},y = {{ - 2} \over 3}$$

$$ \Rightarrow Q\left( {{4 \over 3},{{ - 2} \over 3},0} \right)$$

Vector parallel to the line of intersection

$$\overrightarrow a = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 1} & 2 \cr 2 & 1 & { - 1} \cr } } \right| = - \widehat i + 5\widehat j + 3\widehat k$$

Equation of Line of intersection

$${{x - {4 \over 3}} \over { - 1}} = {{y + {2 \over 3}} \over 5} = {{z - 0} \over 3} = \lambda $$ (say)

Let coordinates of foot of perpendicular be

$$F\left( { - \lambda + {4 \over 3},5\lambda - {2 \over 3},3\lambda } \right)$$

$$\overrightarrow {PF} = \left( { - \lambda + {1 \over 3}} \right)\widehat i + \left( {5\lambda - {8 \over 3}} \right)\widehat j + (3\lambda )\widehat k$$

$$\overrightarrow {PF} .\overrightarrow a = 0$$

$$ \Rightarrow \lambda - {1 \over 3} + 25\lambda {{ - 40} \over 3} + 9\lambda = 0$$

$$ \Rightarrow 35\lambda = {{41} \over 3} \Rightarrow \lambda = {{41} \over {105}}$$

Now, $$\alpha = - \lambda + {4 \over 3},\beta = 5\lambda - {2 \over 3},\gamma = 3\lambda $$

$$ \Rightarrow \alpha + \beta + \gamma = 7\lambda + {2 \over 3}$$

$$ = 7\left( {{{41} \over {105}}} \right) + {2 \over 3}$$

$$ = {{51} \over {15}}$$

$$ \Rightarrow 35(\alpha + \beta + \gamma ) = {{51} \over {15}} \times 35 = 119$$

3d Geometry

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