Consider the lines L _ 1 : (x - 1)/(2) = (y)/(- 1) = (z + 3)/(1) ,...

Given equation of two line L₁ and L₂

${L}_{1}\text{:}\dfrac{x - 1}{2} = \dfrac{y}{- 1} = \dfrac{z + 3}{1} = {k}_{1}\left(say\right){L}_{2}: \dfrac{x - 4}{1} = \dfrac{y + 3}{1} = \dfrac{z + 3}{2} = {k}_{2}\left(say\right)$

And equation of two planes ${P}_{1}$ and ${P}_{2}$

${P}_{1}\text{:}7x + y + 2z = 3\text{,}{P}_{2}\text{:}3x + 5y - 6z = 4\text{.}$

Evaluating unit vector \overset{^}{n} perpendicular to both the given planes P₁ and P₂
⇒ \overset{\to}{n} = \left|\begin{cases}\overset{^}{\text{i}} & \overset{^}{\text{j}} & \overset{^}{k} \\ 7 & 1 & 2 \\ 3 & 5 & - 6\end{cases}\right| = \overset{^}{i}\left(\left(- 6 . 1\right) - \left(2 . 5\right)\right) - \overset{^}{j}\left(\left(- 6 . 7\right) - \left(2 . 3\right)\right) + \overset{^}{k}\left(\left(5 . 7\right) - \left(3 . 1\right)\right) = \left(- 6 - 10\right)\overset{^}{i} - \left(- 42 - 6\right)\overset{^}{j} + \left(35 - 3\right)\overset{^}{k} = - 16\overset{^}{i} + 48\overset{^}{j} + 32\overset{^}{k} ⇒ \overset{\to}{n} = \left(\overset{^}{i} - 3\overset{^}{j} - 2\overset{^}{k}\right)

Now, the point of intersection of line${L}_{1}$ and ${L}_{2}$be $( {x}_{1}, {y}_{1}, {z}_{1} )$

So, the point will satisfy the equation of line

Hence, $\dfrac{{x}_{1} - 1}{2} = \dfrac{{y}_{1}}{- 1} = \dfrac{{z}_{1} + 3}{1} = {k}_{1} ⇒ \dfrac{{x}_{1} - 1}{2} = {k}_{1} ⇒ {x}_{1} = 2{k}_{1} + 1 . . . . . \left(i\right)also, L2: \dfrac{{x}_{1} - 4}{1} = \dfrac{{y}_{1} - 3}{1} = \dfrac{{z}_{1} + 3}{2} = {k}_{2} ⇒ \dfrac{{x}_{1} - 4}{1} = {k}_{2} ⇒ {x}_{1} = {k}_{2} + 4 . . . . . \left(ii\right)$

Comparing equation (i) and (ii) we get

$⇒ {k}_{2} + 4 = 2{k}_{1} + 1 ⇒ {k}_{2} = 2{k}_{1} - 3 . . . . . \left(a\right)$

Similarly,

$\dfrac{{y}_{1}}{- 1} = {k}_{1} ⇒ {y}_{1} = - {k}_{1} . . . \left(iii\right)\dfrac{{y}_{1} + 3}{1} = {k}_{2} ⇒ {y}_{1} = {k}_{2} - 3 . . . \left(iv\right)$

Compare equation (iii) and (iv)

-k₁=k₂-3

k₂=3-k₁ .... (b)

Now, compare equation (a) and (b)

$3 - {k}_{1} = 2{k}_{1} - 3 ⇒ 3{k}_{1} = 6 ⇒ {k}_{1} = 2and{k}_{2} = 3 - 2 = 1$

So, point of intersection is obtained by substituting k₁ in L₁ or k₂ in L₂

So,

$\dfrac{{x}_{1} - 2}{2} = \dfrac{y}{- 1} = \dfrac{z + 3}{1} = 2$

Now, equation of plane having unit vector

\overset{⇀}{A} = \overset{^}{i} - 3\overset{^}{j} - 2\overset{^}{k} and passing through point (5,-2,-1)

$⇒ 1\left(x - 5\right) - 3\left(y + 2\right) - 2\left(z + 1\right) = 0 ⇒ x - 5 - 3y - 6 - 2z - 2 = 0 ⇒ x - 3y - 2z = 13$

Compare the above equation with ax+by+cz=d

So, $a = 1, b = - 3, c = - 2, d = 13$