UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The image of a point in a plane lies on the line through the point perpendicular to the plane. The midpoint of the point and its image lies on the given plane.Explanation:Let the image be Q = $(x, y, z)$.The normal to the plane $3x - 2y + 2z + 1 = 0$ has direction ratios $(3, -2, 2)$.The line through $P(-2, 1, -5)$ parallel to the normal is:$x = -2 + 3t$, $y = 1 - 2t$, $z = -5 + 2t$ (where t is a parameter).The midpoint L of P and Q lies on the plane. Since Q lies on the line, Q = $(-2 + 3t, 1 - 2t, -5 + 2t)$.Then L = $\left(\frac{-2 + (-2+3t)}{2}, \frac{1 + (1-2t)}{2}, \frac{-5 + (-5+2t)}{2}\right)$ = $\left(\frac{3t}{2} - 2,\ 1 - t,\ t - 5\right)$.Substitute L into the plane equation:$3\left(\frac{3t}{2} - 2\right) - 2(1 - t) + 2(t - 5) + 1 = 0$.Simplify: $\frac{9t}{2} - 6 - 2 + 2t + 2t - 10 + 1 = 0 \rightarrow \frac{9t}{2} + 4t - 17 = 0 \rightarrow \frac{17t}{2} = 17 \rightarrow t = 2$.Thus Q = $(-2 + 6, 1 - 4, -5 + 4) = (4, -3, -1)$. Statement 1 is correct.Length PQ = $\sqrt{(4+2)^2 + (-3-1)^2 + (-1+5)^2} = \sqrt{36 + 16 + 16} = \sqrt{68} \approx 8.246$ units. Since $8.246 > 8$, statement 2 is correct.Midpoint L when $t = 2$ is $(1, -1, -3)$. Checking the plane: $3(1) - 2(-1) + 2(-3) + 1 = 3 + 2 - 6 + 1 = 0$. Hence L lies on the plane, so statement 3 is correct.Answer:D. 1, 2 and 3.