Answer :

The two given lines are:

Let a, b, c be the direction Ratios of the normal to the plane containing the line (1).

Also, as plane contains line (1), so it passes through (1, 4, 4)

Therefore, equation of plane is -

a (x –1) + b(y – 4)+c(z – 4) = 0 …(3)

We know that if a_{1}, b_{1}, c_{1} are direction Ratios of a plane and a_{2}, b_{2}, c_{2} are direction Ratios of a line parallel to it then,

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

As, required plane is parallel to line (1),

⇒ 3a+2b – 2c = 0 … (4)

Also, as required plane is parallel to line (2),

⇒ 2a – 4b + 1c = 0 … (5)

Solving (4) and (5),

Let,

Putting, a = 6λ, b = λ7, c = 16λ in (3), we get

⇒ 6λ(x –1) + 7λ(y – 4) + 16λ(z – 4) = 0

⇒ λ((6x –6) + (7y – 28) + (16z – 64)) = 0

⇒ 6x + 7y +16z – 6 – 28 – 64 = 0,

⇒ 6x + 7y +16z – 98 = 0,

which is the required equation of the plane.

Since line (2) is parallel to required plane

∴ SD between two lines = Perpendicular distance of the point

(–1, 1 –2) from the plane.

Distance between a point (x’, y’, z’) from a line

ax + by + cz – d = 0 is -

Therefore,

**OR**

Equation of given line,

Also, (1) is coplanar with the line determined by the planes

x + 2y + 3z – 8 = 0 … (2)

and

2x + 3y + 4z – 11 = 0 … (3)

So, need to show there exists a plane passing through intersection of planes (2) and (3) containing the line (1).

Any plane passing through intersection of two planes P1 and P2 is

P1 + kP2 = 0

Where k is some constant,

So, equation of the plane passing through the intersection of planes (2) and (3) is -

(x + 2y + 3z – 8) + k (2x + 3y + 4z – 11) = 0 … (4)

As, (4) passes through (-1, -1, -1) so, substituting the coordinates of the point (–1, –1, –1) in (4), we get,

(–1–2–3–8) + k (–2–3–4–11) = 0

⇒ (-14) + k (-20) = 0

Putting in (4) we get,

⇒ 10(x + 2y + 3z – 8) -7(2x + 3y + 4z – 11) = 0

⇒ 10x + 20y + 30z – 80 – 14x + 21y + 28z – 77 = 0

⇒ 4x + y – 2z + 3 = 0

So, plane passing through intersection of planes (2) and (3) is-

4x + y – 2z + 3 =0 … (5)

Now we find value of a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} where a_{1}, b_{1}, c_{1} are

Direction Ratios of the line (1) and a_{2}, b_{2}, c_{2} are Direction Ratios of the normal to the plane (5)

From equation (1) – (a_{1}, b_{1}, c_{1}) = (1, 2, 3) and

From equation (5) – (a_{2}, b_{2}, c_{2}) = (4, 1, -2)

So, a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (1)(4) + (2)(1) + (3) (-2)

= 4 + 2 – 6

= 0

i.e., a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

which implies line (1) lies in plane (5)

Hence the two lines are coplanar and the equation of the plane containing them is 4x + y – 2z + 3 = 0

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