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Hint- Use a family of parallel planes . Equation of all planes parallel to $ax + by + cz + d = 0$ lies on this family $ax + by + cz + e = 0$.

Now, the equation of all planes parallel to $3x + 2y - 2z + 15 = 0$ lies on the family of planes $3x + 2y - 2z + d = 0$.

But we require a unique plane pass through the point $\left( {3,8,2} \right)$. So, we put the point in the family of planes.

$

3 \times 3 + 2 \times 8 - 2 \times 2 + d = 0 \\

\Rightarrow 9 + 16 - 4 + d = 0 \\

\Rightarrow d = - 21 \\

$

So, the required plane is $3x + 2y - 2z - 21 = 0..........\left( 2 \right)$

The required plane also passes through the given line. So, we find an intersection point.

Equation of line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3} = p$

$ \Rightarrow x = 2p + 1,y = 4p + 3,z = 3p + 2$

These points also satisfy the equation of the plane. So,

$

\Rightarrow 3\left( {2p + 1} \right) + 2(4p + 3) - 2(3p + 2) - 21 = 0 \\

\Rightarrow 6p + 3 + 8p + 6 - 6p - 4 - 21 = 0 \\

\Rightarrow 8p = 16 \\

\Rightarrow p = 2 \\

$

Intersection points of the plane and line are $x = 2 \times 2 + 1 = 5$ ,$y = 4 \times 2 + 3 = 11$ , $z = 3 \times 2 + 2 = 8$ .

Coordinate of intersection $\left( {5,11,8} \right)$ .

Distance of point $\left( {3,8,2} \right)$ from the line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3}$ is equal to distance between $\left( {3,8,2} \right)$ and $\left( {5,11,8} \right)$.

Distance $

= \sqrt {{{\left( {5 - 3} \right)}^2} + {{\left( {11 - 8} \right)}^2} + {{\left( {8 - 2} \right)}^2}} \\

\\

$

$

\Rightarrow \sqrt {4 + 9 + 36} \\

\Rightarrow \sqrt {49} = 7 \\

$

So, the correct option is (d).

Note-Whenever we face such types of problems we use some important points. First find the equation of plane passing through the point and parallel to another plane with the help of the family of planes then find the point of intersection of plane and line then using distance formula we get the required answer.

Now, the equation of all planes parallel to $3x + 2y - 2z + 15 = 0$ lies on the family of planes $3x + 2y - 2z + d = 0$.

But we require a unique plane pass through the point $\left( {3,8,2} \right)$. So, we put the point in the family of planes.

$

3 \times 3 + 2 \times 8 - 2 \times 2 + d = 0 \\

\Rightarrow 9 + 16 - 4 + d = 0 \\

\Rightarrow d = - 21 \\

$

So, the required plane is $3x + 2y - 2z - 21 = 0..........\left( 2 \right)$

The required plane also passes through the given line. So, we find an intersection point.

Equation of line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3} = p$

$ \Rightarrow x = 2p + 1,y = 4p + 3,z = 3p + 2$

These points also satisfy the equation of the plane. So,

$

\Rightarrow 3\left( {2p + 1} \right) + 2(4p + 3) - 2(3p + 2) - 21 = 0 \\

\Rightarrow 6p + 3 + 8p + 6 - 6p - 4 - 21 = 0 \\

\Rightarrow 8p = 16 \\

\Rightarrow p = 2 \\

$

Intersection points of the plane and line are $x = 2 \times 2 + 1 = 5$ ,$y = 4 \times 2 + 3 = 11$ , $z = 3 \times 2 + 2 = 8$ .

Coordinate of intersection $\left( {5,11,8} \right)$ .

Distance of point $\left( {3,8,2} \right)$ from the line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3}$ is equal to distance between $\left( {3,8,2} \right)$ and $\left( {5,11,8} \right)$.

Distance $

= \sqrt {{{\left( {5 - 3} \right)}^2} + {{\left( {11 - 8} \right)}^2} + {{\left( {8 - 2} \right)}^2}} \\

\\

$

$

\Rightarrow \sqrt {4 + 9 + 36} \\

\Rightarrow \sqrt {49} = 7 \\

$

So, the correct option is (d).

Note-Whenever we face such types of problems we use some important points. First find the equation of plane passing through the point and parallel to another plane with the help of the family of planes then find the point of intersection of plane and line then using distance formula we get the required answer.

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