Let a line passing through the point $(4,1,0)$ intersect the line $L_1 ; \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $\mathrm{A} \quad(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$.
Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right|$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & L_1=\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p \\ & L_2=\frac{x-6}{1}=\frac{y}{1}=\frac{z-4}{-1}=q \\ & A(2 p+1,3 p+2,4 p+3) \\ & B(q+6, q, 4-q) \\ & \text { D.R. of } P A=2 p-3,3 P+1,4 p+3 \\ & \text { D.R. of } P B=q+2, q-1,4-q \\ & \frac{2 p-3}{q+2}=\frac{3 p+1}{q-1}=\frac{4 p+3}{4-q}\end{aligned}$
$\begin{aligned}
& 2 p q-2 p-3 q+3=3 p q+6 p+q+2 \\ & \mathrm{pq}+\mathrm{rp}+4 \mathrm{q}-1=0 \\ & 12 p-3 p q+4-q=4 p q+3 q-4 p-3 \\ & 7 p q-16 p+4 q-7=0 \\ & 8 p-2 p q-12+3 q=4 p q+8 p+3 q+6 \\ & 6 \mathrm{pq}=-18 \quad \therefore \mathrm{pq}=-3 \\ & 8 p+4 q=4 \quad \Rightarrow 2 p+q=1 \\ & -21-16 p+4 q-7 \quad \Rightarrow 4 p-q=-7 \\ & 16 \mathrm{p}-4 \mathrm{q}=-28 \quad \therefore \mathrm{p}=-1, \mathrm{q}=3 \\ & \mathrm{~A}(-1,-1,-1) \quad \mathrm{B}(9,3,1)
\end{aligned}$
$\left|\begin{array}{ccc}1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1\end{array}\right|=\left|\begin{array}{ccc}0 & -1 & 0 \\ -1 & -1 & -1 \\ 9 & 3 & 1\end{array}\right|=1(-1+9)=8$ *

Let a line passing through the point $(4,1,0)$ intersect the line $L_1 ;…

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