Question

A plane containing the point $(3,2,0)$ and the line $\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}$ also contains the point

• A. $(0,7,-10)$
• B. $(0,7,10)$
• C. $(0,3,1)$
• D. $(0,-3,1)$

Answer 1

The correct option is B $(0,7,10)$

Given that point $(3,2,0)$ lies on plane and line $\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}$ also lies on the plane

$(1,5,4)$ are Drs of given line

The plane contains points $(3,2,0)$ and $(1,2,3)$

The Drs of line joining these two points is $(2,0,-3)$

Therefore Drs of normal to plane is $(i+5j+4k)\times (2i-3k)=-15i+11j-10k$

Therefore the equation of plane is $-15x+11y-10z=k$

Point $(3,2,0)$ lies on plane , so we get $k=-23$

So the equation of plane is $15x-11y+10k=23$

Point $(0,7,10)$ satisfies the plane eqaution

Therefore the correct option is $B$