MathematicsMedium249×since 2002Q2603Let the image of the point P(2,−1,3)P(2,-1,3)P(2,−1,3) in the plane x+2y−z=0x+2 y-z=0x+2y−z=0 be QQQ. Then the distance of the plane 3x+2y+z+29=03 x+2 y+z+29=03x+2y+z+29=0 from the point QQQ is :A2142\sqrt{14}214B2227\frac{22\sqrt2}{7}7222C2427\frac{24\sqrt2}{7}7242D3143\sqrt{14}314Check answerSkip