A) \[2x-3y+6z+25=0\]
B) \[3x-2y+6z+25=0\]
C) \[2x-3y+6z-25=0\]
D) \[3x-2y+6z-25=0\]
Correct Answer: C
Solution :
Since, plane is parallel to a given vector, it implies that normal of plane must perpendicular to the given vectors. Given point to which plane passes through is (2, -1, 3). Let A, B and C are direction ratios of its normal. \[\therefore \]Equation of plane is \[A(x-2)+B(y+1)+C(z-3)=0\] Now, normal to plane\[A\mathbf{i}+B\mathbf{j}+C\mathbf{k}\]is perpendicular to the given vectors \[\mathbf{a}=3\mathbf{i}+0\mathbf{j}-\mathbf{k}\] and \[\mathbf{b}=-3\mathbf{i}+2\mathbf{j}+2\mathbf{k}\] \[\therefore \] \[3A+0B-C=0\] ... (i) and \[-3A+2B+2C=0\] ... (ii) From Eqs. (i) and (ii), we get \[\frac{A}{2}=\frac{B}{-3}=\frac{C}{6}\] \[\therefore \]Equation of plane be \[2(x-2)-3(y+1)+6(z-3)=0\] i.e., \[2x-3y+6z-25=0\]
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