A plane P contains the line of intersection of the plane $\overrightarrow{r} .( \^i+\^j+\^k )=6$ and $\overrightarrow{r }.( 2\^i+3\^j+4\^k) = − 5$ . If P passes through the point (0, 2, –2), then the square of distance of the point (12, 12, 18) from the plane P is

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A plane P contains the line of intersection of the plane   $\overrightarrow{r} .( \^i+\^j+\^k )=6$  and  $\overrightarrow{r }.( 2\^i+3\^j+4\^k) = − 5$ .  If P passes through the point (0, 2, –2), then the square of distance of the point (12, 12, 18) from the plane P is

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The equation of the plane is obtained by solving the line of intersection of the given planes: $$ (x + y + z - 6) + \lambda(2x + 3y + 4z + 5) = 0 $$ Passing through the point (0, 2, -2), we solve for $\lambda$: $$ (-6) + \lambda(6 - 8 + 5) = 0 \implies \lambda = 2 $$ Thus, the equation of the plane is: $$ 5x + 7y + 9z + 4 = 0 $$ The distance from the point (12, 12, 18) to the plane is given by: $$ d = \frac{|60 + 84 + 162 + 4|}{\sqrt{25 + 49 + 81}} = \frac{310}{\sqrt{155}} $$ Squaring the distance: $$ d^2 = 310 \times 310 = 620 $$

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A plane P contains the line of intersection of the plane \(\overrightarrow{r} .( \^i+\^j+\^k )=6\) and \(\overrightarrow{r }.( 2\^i+3\^j+4\^k) = − 5\). If P passes through the point (0, 2, –2), then the square of distance of the point (12, 12, 18) from the plane P is

The Correct Option is C

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