Question

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 

• A. 155
• B. 310
• C. 620
• D. 1240

Answer 1

The Correct Option is C

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: \[ d^2 = 310 \times 310 = 620 \]