Consider the lines L ⁡ 1 : x - 1 2 = y - 1 = z + 3 1 , L ⁡ 2 : x - 4 1 = y + 3 1 = z + 3 2 and…

Given equation of two line L₁ and L₂

$$\begin{gathered} L_1\text{:}\frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1}=k_1 \left(say\right) \\ {L}_{2 } : \frac{x-4}{1}=\frac{y+3}{1} =\frac{z+3}{2}=k_2 \left(say\right) \end{gathered}$$

And equation of two planes $P_1$ and $P_2$

$$\begin{gathered} P_1\text{: }7x+y+2z=3\text{,} \\ {P}_2\text{: }3x+5y-6z=4\text{.} \end{gathered}$$

Evaluating unit vector $\hat{n}$ perpendicular to both the given planes P₁ and P₂
$$\begin{gathered} \Rightarrow \overrightarrow{n}=\left|\begin{array}{ccc}\hat{\text{i}}& \hat{\text{j}}& \hat{\mathrm{k}}\\ 7& 1& 2\\ 3& 5& -6\end{array}\right| \\ =\hat{i}\left(\left(-6.1\right)-\left(2.5\right)\right)-\hat{j}\left(\left(-6.7\right)-\left(2.3\right)\right)+\hat{k}\left(\left(5.7\right)-\left(3.1\right)\right) \\ =\left(-6-10\right)\hat{i}-\left(-42-6\right)\hat{j}+\left(35-3\right)\hat{k} \\ =-16\hat{i}+48\hat{j}+32\hat{k} \\ \Rightarrow \overrightarrow{n}=\left(\hat{i}-3\hat{j}-2\hat{k}\right) \end{gathered}$$

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, $$\begin{gathered} \frac{x_1-1}{2}=\frac{y_1}{-1}=\frac{z_1+3}{1}=k_1 \\ \Rightarrow \frac{x_1-1}{2}=k_1 \\ \Rightarrow x_1=2k_1+1 ..... \left(i\right) \\ also, L2 : \frac{x_1-4}{1}=\frac{y_1-3}{1}=\frac{z_1+3}{2}=k_2 \\ \Rightarrow \frac{x_1-4}{1}=k_2 \\ \Rightarrow x_1=k_2+4 ..... \left(ii\right) \end{gathered}$$

Comparing equation (i) and (ii) we get

$$\begin{gathered} \Rightarrow k_2+4=2k_1+1 \\ \Rightarrow k_2=2k_1-3 ..... \left(a\right) \end{gathered}$$

Similarly,

$$\begin{gathered} \frac{y_1}{-1}=k_1 \Rightarrow y_1=-k_1 ... \left(iii\right) \\ \frac{y_1+3}{1}=k_2 \Rightarrow y_1=k_2-3 ... \left(iv\right) \end{gathered}$$

Compare equation (iii) and (iv)

-k₁=k₂-3

k₂=3-k₁  .... (b)

Now, compare equation (a) and (b)

$$\begin{gathered} 3-k_1=2k_1-3 \\ \Rightarrow 3k_1=6 \\ \Rightarrow k_1=2 and k_2=3-2=1 \end{gathered}$$

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

So,

$\frac{x_1-2}{2}=\frac{y}{-1}=\frac{z+3}{1}=2$

Now, equation of plane having unit vector

$\stackrel{\rightharpoonup }{A}=\hat{i}-3\hat{j}-2\hat{k}$ and passing through point (5,-2,-1)

$$\begin{gathered} \Rightarrow 1\left(x-5\right)-3\left(y+2\right)-2\left(z+1\right)=0 \\ \Rightarrow x-5-3y-6-2z-2=0 \\ \Rightarrow x-3y-2z=13 \end{gathered}$$

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

So, $a=1, b=-3, c=-2, d=13$