Let the equation of plane be ax + by + cz + d = 0 …. (1)
Since the plane passes through the point A (0, 0, 0) and B(3, -1, 2), we have a × 0 + b × 0 + c × 0 + d = 0
⇒ d = 0 …. (2)
Similarly for point B (3, -1 , 2), a x 3 + b x (-1) + c x 2 + d = 0
3a – b + 2c = 0 (Using, d = 0) …(3)
Given equation of the line is
$\frac{\mathit{x}-4}{1}$ = $\frac{\mathit{y}+3}{-4}$ = $\frac{\mathit{z}-1}{7}$
We can also write the above equation as
$\frac{\mathit{x}-4}{1}$ = $\frac{\mathit{y}-\left(-3\right)}{-4}$ = $\frac{\mathit{x}-\left(-1\right)}{7}$
The required plane is parallel to the above line.
Therefore, a × 1 + b × (-4) + c × 7= 0
⇒ a – 4b + 7c = 0 … (4)
Cross multiplying equations (3) and (4), we obtain:
$\frac{\mathit{a}}{\left(-1\right)\times 7-\left(-4\right)\times 2}$ = $\frac{\mathit{b}}{2\times 1-3\times 7}$ = $\frac{\mathit{c}}{3\times \left(-4\right)-1\times \left(-1\right)}$
⇒ $\frac{\mathit{a}}{-7+8}$ = $\frac{\mathit{b}}{2-21}$ = $\frac{\mathit{c}}{-12+1}$
⇒ $\frac{\mathit{a}}{1}$ = $\frac{\mathit{b}}{-19}$ = $\frac{\mathit{c}}{-11}$ = k
⇒ a = k , b = - 19k , c = - 11k
Substituting the values of a, b and c in equation (1), we obtain the equation of plane as:
kx - 19ky - 11kz + d = 0
⇒ k (x - 19y - 11z) = 0 (From equation(2))
⇒ x - 19y - 11z = 0
So, the equation of the required plane is x – 19y - 11z = 0