Statement-1 : Let A, B, C be the images of point P(a, b, c) in YZ, ZX and XY planes respectively. Then, the equation of the plane passing through points A, B, C cuts intercepts a, b, c on the coordinates axes.
Statement-2: The image $(α,β,γ )$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0 $ is given by
$\frac{\alpha - x_1}{a}=\frac{\beta - y_1}{b}=\frac{\gamma -z_1}{c}=-\frac{2(ax_1 +by_1 +cz_1+d)}{a^2+b^2+c^2}$
Answer & explanation
Correct answer: option 1
Clearly statement-2 is true .
Using statement-2, the images of point P (a, b, c) in yz, zx and xy-planes are A(-a, b, c), B(a, -b, c( and C(a, b, -c) respectively.
Let the equation of the plane passing through (-a, b, c) be
$U(x + a) + V(y - b) + W(z-c) = 0 $
It passes through (a, -b, c) and (a, b, -c).
$∴ 2Ua - 2 Vb + 0W= 0 $
$2 Ua + 0V - 2 Wc = 0 $
$⇒ \frac{U}{4bc}=\frac{V}{4ac}=\frac{W}{4ab}$
Substituting the values of U, V, W in (i), we obtain
$bc ( x + a) + ca (y- b) + ab (z-c) = 0 ⇒ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}= 1 $
Clearly, this plane cuts intercepts a, b and c on the coordinate axes.