A) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-14=0\]
B) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+x+2y+3z=0\]
C) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-x-2y-3z=0\]
D) None of these
Correct Answer: C
Solution :
[c] Let \[P(\alpha ,\beta ,\gamma )\] be the foot of the perpendicular from the origin \[O(0,0,0)\] to the plane so, the plane passes through \[P(\alpha ,\beta ,\gamma )\] and is perpendicular to OP. clearly direction ratios of OP. i.e,. normal t the plane are \[\alpha ,\beta ,\gamma \] Therefore, equation of the plane is \[\alpha (x-\alpha )+\beta (y-\beta )+\gamma (z-\gamma )=0\] This plane passes though the fixed point\[(1,2,3)\], so \[\alpha (1-\alpha )+\beta (2-\beta )+\gamma (3-\gamma )=0\] or \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}-\alpha -2\beta -3\gamma =0\] Generalizing \[\alpha ,\beta \] and \[\gamma ,\] locus of \[P(\alpha ,\beta ,\gamma )\] is \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-x-2y-3z=0\]
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