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JEE MATHEMATICS MCQ: 3D PLANES

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Application +4 | -1
Through the point P(h,k,l)P(h, k, l) a plane is drawn at right angles to OPO P to meet co-ordinate axes at A,BA, B and CC. If OP=pO P=p then the area of the ā–³ABC\triangle A B C is
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Correct option: A
p52hkl\frac{p^5}{2 h k l}
Option A (p52hkl\frac{{p^5}}{{2 h k l}}) is correct because it represents the area of the triangle ABC, derived from the equation of the plane and the intercepts on the coordinate axes.

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Let xāˆ’23=y+1āˆ’2=z+3āˆ’1\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1} lie on the plane pxāˆ’qy+p x-q y+ z=5z=5, for some p,q∈Rp, q \in \mathbb{R}. The shortest distance of the plane from the origin is : If the planes xāˆ’cyāˆ’bz=0,cxāˆ’y+az=0x-c y-b z=0, c x-y+a z=0 and bx+ayāˆ’z=0b x+a y-z=0 pass through a straight line, then the value of a2+b2+c2+2abca^2+b^2+c^2+2 a b c is: Let the plane 2xāˆ’3y+9z=02 x-3 y+9 z=0 is PP, the equation of line passing through (0,1,āˆ’1)(0,1,-1) and lying-in plane 4xāˆ’2y+7z+9=04 x-2 y+7 z+9=0 & parallel to PP is: A plane which passes through the point (3,2,0)(3,2,0) and the line xāˆ’41=yāˆ’75=zāˆ’44\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4} is The plane ax+a x+ by =0=0 is rotated through an angle α\alpha about its line of intersection with the plane z=0z=0. If the equation of plane in new position is ax+by±za2+b2tan⁔α+c=0\mathrm{ax}+\mathrm{by} \pm \mathrm{z} \sqrt{\mathrm{a}^2+\mathrm{b}^2} \tan \alpha+\mathrm{c}=0. Find c. A variable plane passes through a fixed point (3,2,1)(3,2,1) and meets x,yx, y and zz axes at A,BA, B and CC respectively. A plane is drawn parallel to yzy z-plane through AA, a second plane is drawn parallel zxz x-plane through BB and a third plane is drawn parallel to xyx y-plane through CC. Then the locus of the point of intersection of these three planes, is
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