# Find the equation

We know that the equation of plane passing through (x1,y1,z1) is given by

a(x – x1) + b(y – y1) + c(z – z1) = 0 …… (1)

So, equation of plane passing through (3,4,1) is

a(x – 3) + b(y – 4) + c(z – 1) = 0 ……(2)

It also passes through (0,1,0)

So, equation (2) must satisfy the point (0,1,0)

a(0 – 3) + b(1 – 4) + c(0 – 1) = 0

– 3a – 3b – c = 0

3a + 3b + c = 0 ……(3)

We know that line is parallel to plane a2x + b2y + c2z + d2 = 0 if a1a2 + b1b2 + c1c2 = 0 …… (4)

So,

a×2 + b×7 + c×5 = 0

2a + 7b + 5c = 0 ……(5)

Solving equation (3) and (5) by cross multiplication we have,

a = 8k, b = – – 13k and c = 15k

Putting the value in equation (2)

8k(x – 3) – 13k(y – 4) + 15k(z – 1) = 0

8kx – 24k – 13ky + 52k + 15kz – 15k = 0

Dividing by k we have

8x – 13y + 15z + 13 = 0

Equation of required plane is 8x – 13y + 15z + 13 = 0

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