QJEEAdv16Ma1127: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Ma1127 Solution: Let be the image of the point with respect to the plane Then…

Let P(x₁, y₁, z₁) image of Q(3, 1, 7) w.r.t. the plane x $-$ y + z = 3.

Let R be the point on plane which is midpoint of the line joining P and Q.

The equation of the line PQ is

${{x - 3} \over 1} = {{y - 1} \over { - 1}} = {{z - 7} \over 1} = \lambda$

Therefore, (x, y, z) = (3 + $\lambda$ , 1 $-$ $\lambda$ , 7 + $\lambda$ ) lies on plane.

$3 + \lambda - 1 + \lambda + 7 + \lambda = 3$

$3\lambda + 6 = 0 \Rightarrow \lambda = - 2$

The point R is (3 $-$ 2, 3, 7 $-$ 2) = (1, 3, 5).

Now, ${{{x_1} + 3} \over 2} = 1 \Rightarrow {x_1} = - 1$

${{{y_1} + 1} \over 2} = 3 \Rightarrow {y_1} = 5$

${{{z_1} + 7} \over 2} = 5 \Rightarrow {z_1} = + 3$

That is, the point P is P( $-$ 1, 5, +3).

Now, the equation of the plane passing through P is

$a(x + 1) + b(y - 5) + c(z - 3) = 0$

This plane contains the line:

${x \over 1} = {y \over 2} = {z \over 1}$

This is, $a(0 + 1) + b(0 - 5) + c(0 - 3) = 0 \Rightarrow a = 5b + 3c$

$a + 2b + c = 0 \Rightarrow 7b + 4c = 0 \Rightarrow b = {{ - 4c} \over 7}$

$a = - {{20c} \over 7} + 3c = {c \over 7}$

Now, the equation of the plane is obtained as follows:

${c \over 7}(x + 1) - {{4c} \over 7}(y - 5) + c(z - 3) = 0$

$(x + 1) - 4(y - 5) + 7(z - 3) = 0$

$x + 1 - 4y + 20 + 7z - 21 = 0$

$x - 4y + 7z = 0$