If the plane $$2ax-3ay+4az+6=0$$ passes through the midpoint of the line joining the centres of the spheres $${x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13$$ and $${x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8$$ then $$a$$ equals :
A.
$$-1$$
B.
$$1$$
C.
$$-2$$
D.
$$2$$
Answer :
$$-2$$
Solution :
Centers of given spheres are $$\left( { - 3,\,4,\,1} \right)$$ and $$\left( {5,\, - 2,\,1} \right).$$
Mid point of centres is (1, 1, 1).
Satisfying this in the equation of plane, we get
$$\eqalign{
& 2a - 3a + 4a + 6 = 0 \cr
& \Rightarrow a = - 2 \cr} $$
Releted MCQ Question on Geometry >> Three Dimensional Geometry
Releted Question 1
The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :
If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :