Question

The line joining the points $(1,1,2)$ and $(3,-2,1)$ meets the plane $3x+2y+z=6$ at the point.

• A. $(1,1,2)$
• B. $(3,-2,1)$
• C. $(2,-3,1)$
• D. $(3,2,1)$

Answer 1

Answer: (B)

$(3,-2,1)$

The straight line joining the points $(1,1,2)$ and $(3,-2,1)$ is $\frac{x-1}{2}=\frac{y-1}{-3}=\frac{z-2}{-1}=r\left(\text{say}\right)$
Therefore, point is $(2r+1,1-3r,2-r)$ which lies on $3x+2y+z=6$.
Thus $3(2r+1)+2(1-3r)+2-r=6$
$\Rightarrow r=1$
So, required point is $(3,-2,1)$.