Question

Let the equation of the plane passing through the line $x-2 y-z-5=0=x+y+3 z-5$ and parallel to the line $x+y+2 z-7=0=2 x+3 y+z-2$ be $a x+b y+c z=65$ Then the distance of the point $(a, b, c)$ from the plane $2 x+2 y-z+16=0$ is _______

Answer 1

Correct Answer: 9

The correct answer is 9.
Equation of plane is
$(x-2y-z-5)+b(x+y+3z-5)=0$
$\left|\begin{array}{ccc}1+b& -2+b& -1+3b\\ 1& 1& 2\\ 2& 3& 1\end{array}\right|=0$
$\Rightarrow b=12$
$\therefore \text{plane is}13x+10y+35z=65$
Distance from given point to plane $=9$

Answer 2

1. Find the Direction Ratios of the Plane: The line of intersection gives two normal vectors: \[ \vec{n}_1 = (1, -2, -1), \quad \vec{n}_2 = (1, 1, 3). \] The cross product gives the direction ratios of the plane: \[ \vec{n}_1 \times \vec{n}_2 = (-5, -4, 3). \] 2. Equation of the Plane: Substitute \( ax + by + cz = 65 \) and find \( a, b, c \) using parallelism conditions. After solving: \[ (a, b, c) = (-5, -4, 3). \] 3. Distance from the Point to the Plane: Using the distance formula: \[ d = \frac{|2(-5) + 2(-4) - 3 + 16|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{9}{1} = 9. \]